How to assign a multiple line value to a bash variable

Don't indent the lines or you'll get extra spaces. Use quotes when you expand "$FOO" to ensure the newlines are preserved.

$ FOO="This is line 1 
This is line 2   
This is line 3"
$ echo "$FOO"
This is line 1
This is line 2
This is line 3

Another way is to use \n escape sequences. They're interpreted inside of $'...' strings.

$ FOO=$'This is line 1\nThis is line 2\nThis is line 3'
$ echo "$FOO"

A third way is to store the characters \ and n, and then have echo -e interpret the escape sequences. It's a subtle difference. The important part is that \n isn't interpreted inside of regular quotes.

$ FOO='This is line 1\nThis is line 2\nThis is line 3'
$ echo -e "$FOO"
This is line 1
This is line 2
This is line 3

You can see the distinction I'm making if you remove the -e option and have echo print the raw string without interpreting anything.

$ echo "$FOO"
This is line 1\nThis is line 2\nThis is line 3

When you initialize FOO you should use line breaks: \n.

FOO="This is line 1\nThis is line 2\nThis is line 3"

Then use echo -e to output FOO.

It is important to note that \n inside "..." is NOT a line break, but literal \ , followed by literal n. It is only when interpreted by echo -e that this literal sequence is converted to a newline character. — wise words from mklement0

---

#!/bin/bash

FOO="This is line 1\nThis is line 2\nThis is line 3"
echo -e $FOO

Output:
This is line 1
This is line 2
This is line 3