Haskell, many previous byte counts 127 * 0.9 = 114.3 bytes

f#(a:b)=f a:f#b;f#x=x
(f&x)0=x;(f&x)i=f$f&x$i-1
i=id
r x=i%(1,x)
(g?x)(a:b)=g(g?x$b)a;(g?x)y=x
f%(a,b)|a>b=[]|1<2=f a:f%(a+1,b)

No loops, just recursion.

# is map: (*2) # [1,2,3] -> [2,4,6]

& is nest: ((*2) & 3) 4 -> 48

i is apply: i (*2) 7 -> 14

r is range: r 4 -> [1,2,3,4]

? is fold: ((+) ? 0) [1,2,3,4] -> 10

% is table: (*2) % (2,4) -> [4,6,8]

As requested an ungolfed version with comments. Note, & and ? are ternary infix operators, which require additional parentheses when called or pattern matched.

f # (a:b) = f a : f#b        -- map on a list (a->head, b->tail) is f a in front of mapping f to b
f # x     = x                -- map on the empty list is the empty list
                             -- (non empty lists are caught in the line before) 

(f & x) 0 = x                -- nesting zero times is x
(f & x) i = f $ f&x $ i-1    -- nesting i times is f (nesting one time less)

i=id                         -- apply in Haskell is just the identity function 

r x = i % (1,x)              -- defined via the "table" of the identity function from 1 to x

(g ? x) (a:b) = g (g?x$b) a  -- folding g into a list (a->head, b->tail) is g applied to (folding g into b) and a
(g ? x) y     = x             -- folding the empty list is x
                             --  again, y must be the empty list, else it would have been handled by the previous line

f % (a,b)                    
  |a>b       = []                -- if iMin is greater than iMax, the table is empty
  |otherwise = f a : f%(a+1,b)   --  otherwise f a in front of the table with iMin increased by one

Thanks to @dfeuer and @Zgarb for some useful hints

Python 2, 305.1 bytes (-10% 376 369 366 349 339 bytes)

exec'e=eval;q=len;m=@,l:e("["+"f(l.pop()),"*q(l)+"][::-1]");n=@,x,l:e("f("*l+"*x"+")"*l);r=@:e("r(f-1)+"*(f>1)+"[f]");a=@,a:e("f(a["+`r(q(a))`[1:-1]$",","-1],a[")+"-1])");f=@,x,l:e("f("*q(l)+"x,["+`r(q(l))`[1:-1]$",","-1]),l[")+"-1])");t=@,n,x:e("[f("+`r(x)[n-1:]`$",","),f(")[1:-1]+")]")'.replace("@","lambda f").replace("$",".replace(")

When expanded, equivalent to:

e=eval;q=len
m=lambda f,l:e("["+"f(l.pop()),"*q(l)+"][::-1]")
n=lambda f,x,l:e("f("*l+"*x"+")"*l)
r=lambda i:e("r(i-1)+"*(i>1)+"[i]")
a=lambda f,a:e("f(a["+`r(q(a))`[1:-1].replace(",","-1],a[")+"-1])")
f=lambda f,x,l:e("f("*q(l)+"x,["+`r(q(l))`[1:-1].replace(",","-1]),l[")+"-1])")
t=lambda f,n,x:e("[f("+`r(x)[n-1:]`.replace(",","),f(")[1:-1]+")]")

No loops!

Well, it does a lot of evaling and if your boss can't stand loops, then they'll HATE eval. But, they're going to have to put up with it

A way to do range in a lambda is appreciated so I don't have to do any functions (Shudder.).

Explanations:

• m=lambda f,l:eval("["+"f(l.pop()),"*len(l)+"][::-1]")
  • Create a string that pops elements from the list, wrap it into a list, reverse it and finally eval it!
• n=lambda f,x,l:eval("f("*l+"*x"+")"*l)
  • Manually create the string with the nesting, and eval it!
• r=lambda i:e("r(i-1)+"*(i>1)+"[i]")
  • Create a string that when evaled, either returns [0] or uses recursion to get the previous results and adds the current index to the list. Evals it.
• a=lambda f,a:eval("f(a["+r(len(a))[1:-1].replace(",","-1],a[")+"-1])")
  • Uses the range function to get the indexes 1-len(list). Replaces the commas in the list stringified with a way to get the correct index of the list a. Evals it!
• f=lambda f,x,l:eval("f("*len(l)+"x,["+r(len(l))[1:-1].replace(",","-1]),l[")+"-1])")
  • Same as apply except replaces the commas with closing brackets, commas and starting the list index.
• t=lambda f,n,x:eval("[f("+r(x)[n-1:].replace(",","),f(")[1:-1]+")]")
  • Same as apply and fold except replaces with ending the function and calling the new one. Evals it!

Map, nest, range, apply, fold, table.

Thanks @Zgarb for a lambda for range!

Javascript ES6, 197 * 0.9 = 177.3 bytes

M=(f,l)=>F((a,b)=>[...a,f(b)],[],l)
N=(f,x,n)=>f(--n?N(f,x,n):x)
A=(f,l)=>f(...l)
R=n=>n--?[...R(n),n+1]:[]
F=(f,x,l,n=l.length)=>n--?f(F(f,x,l,n),l[n]):x
T=(f,i)=>([n,x]=i,M(q=>f(q+n-1),R(x-n+1)))

Map (M=(f,l)=>F((a,b)=>[...a,f(b)],[],l)):

Uses Fold to concat the results of f applied to every member of l onto an empty list. Using built-in functions reduces the this to M=(f,l)=>l.map(f) (didn't use it because it seems cheap...?).

Nest (N=(f,x,n)=>f(--n?N(f,x,n):x)):

Apply f recursively until n is decremented to 0.

Apply (A=(f,l)=>f(...l)):

Uses the spread (...) operator to apply l onto f.

Range (R=n=>n--?[...R(n),n+1]:[]):

Concat n to recursive call of Range until n is decremented to 0.

Fold (F=(f,x,l,n=l.length)=>n--?f(F(f,x,l,n),l[n]):x):

Applies the recursive call of Fold and the n'th element of l to f until n is decremented to 0. Using built-in functions reduces this to F=(f,x,l)=>l.reduce(f,x) (again, seemed cheap...).

Table (T=(f,i)=>([n,x]=i,M(q=>f(q+n-1),R(x-n+1)))):

First initializes n and x to iMin and iMax using destructuring ([n,x]=i), then uses Range to construct the table of values from iMin to iMax. f is then applied over the table using Map and the result is returned.

Julia, 181 bytes

No bonus for me; I used loops liberally. Sorry boss, but loops in Julia are efficient!

M(f,x)=[f(i...)for i=x]
N(f,x,n)=(for i=1:n x=f(x...)end;x)
A(f,x)=f(x...)
R(n)=(i=0;x={};while i<n push!(x,i+=1)end;x)
F(f,x,a)=(for b=a x=f(x,b)end;x)
T(f,i)=[f(j)for j=i[1]:i[2]]

Adding the ellipses after an argument to a function breaks an array, tuple, or what have you into regular function arguments. Otherwise the function will think you're trying to pass an array (or tuple, etc.). It has no effect for single arguments.

Function names:

• Map: M
• Nest: N
• Apply: A
• Range: R
• Fold: F
• Table: T

Python 2.x: 450.6 bytes (493 bytes before 10% discount)

Golfed answer:

y=len
z=lambda a,b:a.append(b)
_=lambda a:a if a is not None else[]
def M(a,b,c=None):
 c=_(c);d=y(b)
 if d:z(c,A(a,b[0]))
 return M(a,b[1:],c)if d else c
def N(a,b,c):d=A(a,b);return N(a,d,c-1)if c>1 else d
A=lambda a,b:a(*b)if type(b)is list else a(b)
def R(a,b=None):b=_(b);b.insert(0,a);return b if a<=1 else R(a-1,b)
def F(a,b,c):d=a(b,c[0]);return F(a,d,c[1:])if y(c)>1 else d
def T(a,b,c=None,d=None):
 if c is None:c=b[0];d=[]
 z(d,a(c));return T(a,b,c+1,d)if c<b[1]else d

This question was fun. I decided to write my functions without using the Python equivalents (though that may have been a valid loophole) and to write the functions as though Python supports tail recursion. To make this work, I used a lot of optional parameters that allow the required calls to still work.

Below I have ungolfed listings for each function.

Apply:

A = lambda function, arguments: function(*arguments) if type(arguments) is list else function(arguments)

Map:

def M(function, arguments, result=None):
    result = result if result is not None else []
    length = len(arguments)
    if length != 0:
        result.append(A(function, arguments[0]))
    return M(function, arguments[1:], result) if length != 0 else result

Nest:

def N(function, arguments, times):
    result = A(function, arguments)
    return N(function, result, times - 1) if times > 1 else result

Note that this function requires that the passed function be able to represent multiple arguments variably. Another approach would have been to enforce that the function always received a single list, but that would have required that the passed functions be able to interpret argument lists. There were assumptions either way, so I chose the one that fit the rest of the system better.

Range:

def R(ceiling, result=None):
    result = result if result is not None else []
    result.insert(0, ceiling)
    return result if ceiling <= 1 else R(ceiling - 1, result)

Fold:

def F(function, initial, rest):
    result = function(initial, rest[0])
    return F(function, result, rest[1:] if len(rest) > 1 else result

Table:

def T(function, iterator, current=None, result=None):
    if current is None:
        current = iterator[0]
        result = []
    result.append(function(current))
    return T(function, iterator, current + 1, result) if current < iterator[1] else result

Here is sample output using the following helper functions:

square = lambda x: x * x
def add(*args):
    addTwo = lambda a, b: a + b
    if len(args) == 1 and type(args[0]) is list:
        return F(addTwo, 0, args[0])
    else:
        return F(addTwo, 0, args)

>>> M(square, R(10))
[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
>>> M(add, [R(i) for i in R(10)])
[1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
>>> T(square, [0, 10])
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
>>> N(square, 2, 4)
65536
>>> N(lambda *args: F(lambda a, b: a * b, 1, args) if len(args) > 1 else str(args[0]) + 'a', R(5), 10)
'120aaaaaaaaa'

Python 3, 218 bytes

The unreadable version:

exec("P!:[f(_)for _ in x];Y!,z:Y(f,f(x),z-1)if z else x;T!:f(*x);H!=0:(H(f-1)if~-f else[])+[f];O!,z:O(f,f(x,z[0]),z[1:])if z else x;N!:(N(f,[x[0],x[1]-1])if x[1]-x[0]else[])+[f(x[1])]".replace("!","=lambda f,x"))

The (more) readable version:

P=lambda f,x:[f(_)for _ in x]
Y=lambda f,x,z:Y(f,f(x),z-1)if z else x
T=lambda f,x:f(*x)
H=lambda f,x=0:(H(f-1)if~-f else[])+[f]
O=lambda f,x,z:O(f,f(x,z[0]),z[1:])if z else x
N=lambda f,x:(N(f,[x[0],x[1]-1])if x[1]-x[0]else[])+[f(x[1])]

Going though it one lambda at a time:

Map function P

P=lambda f,x:[f(_)for _ in x]
Just a simple iterator. Not much to say here.

Nest function Y

Y=lambda f,x,z:Y(f,f(x),z-1)if z else x
Recursing until z hits zero, applying f every time. The if clause at the end feels clunky; perhaps there is a better way to end the recursion.

Apply function T

T=lambda f,x:f(*x)
Python has a nice expansion operator to do all of the heavy lifting for me.

Range function H

H=lambda f,x=0:(H(f-1)if~-f else[])+[f]
This one was harder than I was expecting. Ended up taking a recursive approach. Again, the if-else construction takes up a lot of bytes, and I feel it can be improved. Why does it have a dummy x=0, you ask? It's so that when I compress it with the exec, I can replace =lambda f,x instead of just =lambda f.

Fold function O

O=lambda f,x,z:O(f,f(x,z[0]),z[1:])if z else x
Pretty happy with this one. Just lops off the first element of the array every time it recurses, until there's nothing left.

Table function N

N=lambda f,x:(N(f,[x[0],x[1]-1])if x[1]-x[0]else[])+[f(x[1])]
This one is horrible and I'm sure there's room for improvement. Tried to use the range and map functions previously defined for a map(f,range(x,y)) sort of construction, but without much success. Ended up doing a terrible recursive approach which shares some similarity to the range function.

All the lambdas are wrapped up in an exec with a replace to shorten the byte count significantly.