Jelly, 10 9 8 bytes

Œg+2/€UF

TryItOnline or run all test cases

How?

Œg+2/€UF - Main link: a                 e.g. [2,2,2,4,4,8]
Œg       - group runs of equal elements      [[2,2,2],[4,4],[8]]
   2/€   - pairwise reduce for each with
  +      -     addition                      [[4,2],[8],[8]]
      U  - reverse (vectorises)              [[2,4],[8],[8]]
       F - flatten list                      [2,4,8,8]

Haskell, 47 57 50 bytes

e#l|a:b<-l,e==a= -2*a:b|1<2=e:l
map abs.foldr(#)[]

Uses reduce (or fold as it is called in Haskell, here a right-fold foldr). Usage example: map abs.foldr(#)[] $ [2,2,2,4,4,8] -> [2,4,8,8].

Edit: +10 bytes to make it work for unsorted arrays, too. Merged numbers are inserted as negative values to prevent a second merge. They are corrected by a final map abs.

Brain-Flak 158 96

{({}<>)<>}<>{(({}<>)<><(({})<<>({}<>)>)>)({}[{}]<(())>){((<{}{}>))}{}{{}(<({}{})>)}{}({}<>)<>}<>

Try it online!

Explanation:

1 Reverse the list (moving everything to the other stack, but that doesn't matter)

{({}<>)<>}<>
{        }   #keep moving numbers until you hit the 0s from an empty stack
 ({}<>)      #pop a number and push it on the other stack
       <>    #go back to the original stack
          <> #after everything has moved, switch stacks

2 Do steps 3-6 until there is nothing left on this stack:

{                                                                                         }

3 Duplicate the top two elements (2 3 -> 2 3 2 3)

(({}<>)<><(({})<<>({}<>)>)>)

(({}<>)<>                   #put the top number on the other stack and back on the very top
         <(({})             #put the next number on top after:
               <<>({}<>)>   #copying the original top number back to the first stack
                         )>)

4 Put a 1 on top if the top two are equal, a 0 otherwise (from the wiki)

({}[{}]<(())>){((<{}{}>))}{}

5 If the top two were equal (non-zero on the top) add the next two and push the result

{{}(<({}{})>)}{}
{            }   #skip this if there is a 0 on top
 {}              #pop the 1
   (<      >)    #push a 0 after:
     ({}{})      #pop 2 numbers, add them together and push them back on 
              {} #pop off the 0

6 Move the top element to the other stack

({}<>)<>

7 Switch to the other stack and print implicitly

<>

PHP, 116 Bytes

<?$r=[];for($c=count($a=$_GET[a]);$c-=$x;)array_unshift($r,(1+($x=$a[--$c]==$a[$c-1]))*$a[$c]);echo json_encode($r);

or

<?$r=[];for($c=count($a=$_GET[a]);$c--;)$r[]=$a[$c]==$a[$c-1]?2*$a[$c--]:$a[$c];echo json_encode(array_reverse($r));

-4 Bytes if the output can be an array print_r instead of 'json_encode`

176 Bytes to solve this with a Regex

echo preg_replace_callback("#(\d+)(,\\1)+#",function($m){if(($c=substr_count($m[0],$m[1]))%2)$r=$m[1];$r.=str_repeat(",".$m[1]*2,$c/2);return trim($r,",");},join(",",$_GET[a]));

Perl, 43 + 1 (-p) = 44 bytes

Ton Hospel came up with 41 bytes answer, check it out!

-4 thanks to @Ton Hospel!

Edit : added \b, as without it it was failing on input like 24 4 on which the output would have been 28.

$_=reverse reverse=~s/(\b\d+) \1\b/$1*2/rge

Run with -p flag :

perl -pe '$_=reverse reverse=~s/(\b\d+) \1\b/$1*2/rge' <<< "2 2 2 4 4"

I don't see another way than using reverse twice to right-fold (as just s/(\d+) \1/$1*2/ge would left-fold, i.e 2 2 2 would become 4 2 instead of 2 4). So 14 bytes lost thanks to reverse... Still I think there must be another (better) way (it's perl after all!), let me know if you find it!

Python, 61 bytes

def f(l):b=l[-2:-1]==l[-1:];return l and f(l[:~b])+[l[-1]<<b]

The Boolean b checks whether the last two elements should collapse by checking that they are equal in a way that's safe for lists of length 1 or 0. The last element if then appended with a multiplier of 1 for equal or 2 for unequal. It's appended to the recursive result on the list with that many elements chopped off the end. Thanks to Dennis for 1 byte!

Perl 5.10, 61 50 bytes (49 + 1 for flag)

Thanks to Ton Hospel for saving 11 bytes!

Regex-free solution, with -a flag:

@a=($F[-1]-$b?$b:2*pop@F,@a)while$b=pop@F;say"@a"

Try here!

GNU sed 41 38 37

Includes +1 for -r
-3 Thanks to Digital Trauma
-1 Thanks to seshoumara

:
s,(.*)(1+) \2\b,\1!\2\2!,
t
y,!, ,

Input and output are space separated strings in unary (based on this consensus).

Try it online!

Retina, 32

\d+
$*
r`\b\1 (1+)\b
$1$1
1+
$.&

r on line 3 activates right-to-left regex matching. And this means that the \1 reference needs to come before the (1+) capturing group that it references.

Try it online.

Perl, 41 bytes

Includes +1 for -p

Give input sequence on STDIN:

shift2048.pl <<< "2 2 2 4 4 8 2"

shift2048.pl:

#!/usr/bin/perl -p
s/.*\K\b(\d+) \1\b/2*$1.A/e&&redo;y/A//d

05AB1E, 26 bytes

D¥__X¸«DgL*ê¥X¸«£vy2ôO})í˜

Try it online!

Generalized steps

1. Reduce by subtraction to find where consecutive elements differ
2. Reduce by subtraction over the indices of those places to find the length of consecutive elements
3. Split input into chunks of those lengths
4. Split chunks into pairs
5. Sum each pair
6. Reverse each summed chunk
7. Flatten to 1-dimensional list

JavaScript (ES6), 68 65 58 57 65 bytes

Fixed for unsorted arrays now that it has been clarified that such inputs are to be expected.

f=(a,l=[],m=1)=>(x=a.pop())*m-l?f(a,x).concat(l):x?f(a,2*x,0):[l]

console.log(f([2, 2, 4, 4]));
console.log(f([2, 2, 2, 4, 4, 8]));
console.log(f([2, 2, 2, 2]));
console.log(f([4, 2, 2]));

Alternate version, 72 bytes

A non-recursive version using a regular expression:

a=>a.reverse().join`+`.replace(/(.)\+\1/g,c=>eval(c)).split`+`.reverse()

JavaScript (ES6), 68 bytes

f=a=>a.reduceRight((p,c)=>(t=p[0],p.splice(0,c==t,c==t?c+t:c),p),[])
    
console.log([
  [],
  [2, 2, 4, 4],
  [2, 2, 2, 4, 4, 8],
  [2, 2, 2, 2],
  [4, 4, 2, 8, 8, 2],
  [1024, 1024, 512, 512, 256, 256],
  [3, 3, 3, 1, 1, 7, 5, 5, 5, 5],
].map(f))

Mathematica, 53 bytes

Join@@(Reverse[Plus@@@#~Partition~UpTo@2]&/@Split@#)&

Explanation

Split@#

Split the input into sublists consisting of runs of identical elements. i.e. {2, 2, 2, 4, 8, 8} becomes {{2, 2, 2}, {4}, {8, 8}}.

#~Partition~UpTo@2

Partition each of the sublist into partitions length at most 2. i.e. {{2, 2, 2}, {4}, {8, 8}} becomes {{{2, 2}, {2}}, {{4}}, {{8, 8}}}.

Plus@@@

Total each partition. i.e. {{{2, 2}, {2}}, {{4}}, {{8, 8}}} becomes {{4, 2}, {4}, {16}}.

Reverse

Reverse the results because Mathematica's Partition command goes from left to right, but we want the partitions to be in other direction. i.e. {{4, 2}, {4}, {16}} becomes {{2, 4}, {4}, {16}}.

Join@@

Flatten the result. i.e. {{2, 4}, {4}, {16}} becomes {2, 4, 4, 16}.

Perl, 37 bytes

Includes +4 for -0n

Run with the input as separate lines on STDIN:

perl -M5.010 shift2048.pl
2
2
2
4
4
8
2
^D

shift2048.pl:

#!/usr/bin/perl -0n
s/\b(\d+
)(\1|)$//&&do$0|say$1+$2

Haskell, 56 bytes

g(a:b:r)|a==b=a+b:g r|l<-b:r=a:g l
g x=x
r=reverse
r.g.r

PHP, 86 100 99 bytes

that should do it:

<?$p=array_pop;for($r=[];$v=$p($a=&$_GET[a]);)array_unshift($r,end($a)-$v?$v:2*$p($a));print_r($r);

call in browser with ?a[]=value1&a[]=value2...

Java 7, 133 bytes

Object f(java.util.ArrayList<Long>a){for(int i=a.size();i-->1;)if(a.get(i)==a.get(i-1)){a.remove(i--);a.set(i,a.get(i)*2);}return a;}

Input is an ArrayList, and it just loops backwards, removing and doubling where necessary.

Object f(java.util.ArrayList<Long>a){
    for(int i=a.size();i-->1;)
        if(a.get(i)==a.get(i-1)){
            a.remove(i--);
            a.set(i,a.get(i)*2);
        }
    return a;
}

Julia 205 bytes

t(x)=Val{x}
s(x)=t(x)()
f^::t(1)=f
^{y}(f,::t(y))=x->f(((f^s(y-1))(x)))
g()=[]
g{a}(::t(a))=[a]
g{a}(::t(a),B...)=[a;g(B...)]
g{a}(::t(a),::t(a),B...)=[2a;g(B...)]
K(A)=g(s.(A)...)
H(A)=(K^s(length(A)))(A)

Function to be called is H

eg H([1,2,2,4,8,2,])

This is in no way the shortest way do this in julia. But it is so cool, that I wanted to share it anyway.

• t(a) is a value-type, representing the value (a).
• s(a) is an instance of that value type
• g is a function that dispatches on the difference values (using the value types) and numbers of its parameters. And that is cool
• K just wraps g so that

Extra cool part:

f^::t(1)=f
^{y}(f,::t(y))=x->f(((f^s(y-1))(x)))

This defines the ^ operator to apply to functions. So that K^s(2)(X) is same as K(K(X)) so H is just calling K on K a bunch of times -- enough times to certainly collapse any nested case

This can be done much much shorter, but this way is just so fun.

Python 2, 94 bytes

def f(a,r=[]):
 while a:
    if len(a)>1and a[-1]==a[-2]:a.pop();a[-1]*=2
    r=[a.pop()]+r
 print r

Try it online

PowerShell v2+, 81 bytes

param($n)($b=$n[$n.count..0]-join','-replace'(\d+),\1','($1*2)'|iex)[$b.count..0]

Takes input as an explicit array $n, reverses it $n[$n.count..0], -joins the elements together with a comma, then regex -replaces a matching digit pair with the first element, a *2, and surrounded in parens. Pipes that result (which for input @(2,2,4,4) will look like (4*2),(2*2)) over to iex (short for Invoke-Expression and similar to eval), which converts the multiplication into actual numbers. Stores the resulting array into $b, encapsulates that in parens to place it on the pipeline, then reverses $b with [$b.count..0]. Leaves the resulting elements on the pipeline, and output is implicit.

Test Cases

NB-- In PowerShell, the concept of "returning" an empty array is meaningless -- it's converted to $null as soon as it leaves scope -- and so it is the equivalent of returning nothing, which is what is done here in the first example (after some wickedly verbose errors). Additionally, the output here is space-separated, as that's the default separator for stringified arrays.

PS C:\Tools\Scripts\golfing> @(),@(2,2,4,4),@(2,2,2,4,4,8),@(2,2,2,2),@(4,4,2,8,8,2),@(1024,1024,512,512,256,256),@(3,3,3,1,1,7,5,5,5,5)|%{"$_ --> "+(.\2048-like-array-shift.ps1 $_)}
Invoke-Expression : Cannot bind argument to parameter 'Command' because it is an empty string.
At C:\Tools\Scripts\golfing\2048-like-array-shift.ps1:7 char:67
+   param($n)($b=$n[$n.count..0]-join','-replace'(\d+),\1','($1*2)'|iex)[$b.count. ...
+                                                                   ~~~
    + CategoryInfo          : InvalidData: (:String) [Invoke-Expression], ParameterBindingValidationException
    + FullyQualifiedErrorId : ParameterArgumentValidationErrorEmptyStringNotAllowed,Microsoft.PowerShell.Commands.InvokeExpressionCommand

Cannot index into a null array.
At C:\Tools\Scripts\golfing\2048-like-array-shift.ps1:7 char:13
+   param($n)($b=$n[$n.count..0]-join','-replace'(\d+),\1','($1*2)'|iex)[$b.count. ...
+             ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : InvalidOperation: (:) [], RuntimeException
    + FullyQualifiedErrorId : NullArray

 --> 
2 2 4 4 --> 4 8
2 2 2 4 4 8 --> 2 4 8 8
2 2 2 2 --> 4 4
4 4 2 8 8 2 --> 8 2 16 2
1024 1024 512 512 256 256 --> 2048 1024 512
3 3 3 1 1 7 5 5 5 5 --> 3 6 2 7 10 10

NodeJS - 57 53 bytes

b=a=>{a.reduceRight((z,e)=>{return z+e;});return a;}

Racket 170 bytes

(λ(l)(let g((l(reverse l))(o '()))(cond[(null? l)o][(=(length l)1)(cons(car l)o)]
[(=(car l)(list-ref l 1))(g(drop l 2)(cons(* 2(car l))o))][(g(cdr l)(cons(car l)o))])))

Ungolfed:

(define f1
  (λ (lst)
    (let loop ((lst (reverse lst)) (ol '()))
      (cond
        [(null? lst) ol]
        [(= (length lst) 1)
           (cons (first lst) ol)]
        [(= (first lst) (list-ref lst 1))
           (loop (drop lst 2) (cons (* 2 (first lst)) ol))]
        [(loop (rest lst) (cons (first lst) ol))]
        ))))

Testing:

(f '[])
(f '[2 2 4 4]) 
(f '[2 2 2 4 4 8]) 
(f '[2 2 2 2]) 
(f '[4 4 2 8 8 2])
(f '[1024 1024 512 512 256 256]) 
(f '[3 3 3 1 1 7 5 5 5 5])
(f '[3 3 3 1 1 7 5 5 5 5 5])

Output:

'()
'(4 8)
'(2 4 8 8)
'(4 4)
'(8 2 16 2)
'(2048 1024 512)
'(3 6 2 7 10 10)
'(3 6 2 7 5 10 10)

Julia, 73 82 Bytes

f(l)=l==[]?[]:foldr((x,y)->y[]==x?vcat(2x,y[2:end]):vcat(x,y),[l[end]],l[1:end-1])

Use right fold to build list from back to front (one could also use fold left and reverse the list at the beginning and end).

If the head of the current list is not equal to the next element to prepend, then just prepend it.

Else remove the head of the list (sounds kind of cruel) and prepend the element times 2.

Example

f([3,3,3,1,1,7,5,5,5,5]) 
returns a new list:
[3,6,2,7,10,10]

Java 8, 130 114 bytes

Edit:

• -16 bytes off. Thanks to @Kevin Cruijssen.

Snipet:

s->{int j=0,i=s.length,l[]=new int[i];for(;--i>-1;[j++]=s[i]==s[i-1]?s[i--]*2:s[i]);return Collection.reverse(l);}

Code:

public static int[] shift(int[]a){
  List<int> out = new ArrayList<>();
  for(int i = s.length-1; i>-1;i--){
    if(s[i]==s[i-1]){
      out.add(s[i]*2);
      i--;
    }else{
      out.add(s[i]);
    }
  }
  return Arrays.reverse(out.toArray());
}