How can a Bash script tell how it was run?

From man bash under CONDITIONAL EXPRESSIONS:

-t fd  
    True if file descriptor fd is open and refers to a terminal.

Assuming fd 1 is standard out, if [ -t 1 ]; then should work for you. The Advanced Shell Scripting Guide claims that -t used this way will fail over ssh, and that the test (using stdin, not stdout) should therefore be:

if [[ -t 0 || -p /dev/stdin ]]

-p tests if a file exists and is a named pipe. However, I'd note experientially this is not true for me: -p /dev/stdin fails for both normal terminals and ssh sessions whereas if [ -t 0 ] (or -t 1) works in both cases (see also Gilles comments below about issues in that section of the Advanced Shell Scripting Guide).

If the primary issue is a specialized context from which you wish to call the script to behave in a way appropriate to that context, you can sidestep all these technicalities and save your self some fuss by using a wrapper and a custom variable:

!#/bin/bash

export SPECIAL_CONTEXT=1
/path/to/real/script.sh

Call this live_script.sh or whatever and double click that instead. You could of course accomplish the same thing with command line arguments, but a wrapper would still be needed to make point and click in a GUI file browser work.

Test for the existence of the PS1 environment variable:

# If $PS1 is null, not being run from a shell
if [ -z $PS1 ]

Alternatively you can test for it being not null, i.e., run from the shell

if [ -n $PS1 ]

Another, using the bash options set internal variable, $-.

From .bashrc,

# If not running interactively, don't do anything
case $- in
    *i*) ;;
    *) return;;
esac