Shell – How to assign the cat output of a bash script to a variable in another script

One way to accomplish this is through indirection -- referring to another variable from the value of the first variable.

To demonstrate:

apple="Red"
var="apple"
echo "${!var}"

Results in:

Red

Because bash first takes !var to mean the value of the var variable, which is then interpreted via ${apple} and turned into Red.

As a result, your GetFruitColour.sh script could look like:

#!/bin/bash

source ./fruitcolour.sh

for arg in "$@"
do
  printf 'The colour of %s is %s.\n' "$arg" "${!arg}"
done

I've made the path to the sourced script relative instead of bare, to make it clearer where the file is (if the given filename does not contain a slash, shells will search the $PATH variable for it, which may surprise you).

I've also changed echo to printf.

The functional change is to use the looping variable $arg and the indirect expansion of it to produce the desired values:

$ ./GetFruitColour.sh apple mango
The colour of apple is Red.
The colour of mango is Yellow.

Do note that there's no error-checking here:

$ ./GetFruitColour.sh foo
The colour of foo is .

You may find it easier to use an associative array:

declare -A fruits='([orange]="Orange" [apple]="Red" [mango]="Yellow" [pear]="Green" )'

for arg in "$@"
do
  if [ "${fruits["$arg"]-unset}" = "unset" ]
  then
    echo "I do not know the color of $arg"
  else
    printf 'The colour of %s is %s.\n' "$arg" "${fruits["$arg"]}"
  fi
done