Bash – `set -e` inside a bash function

Your script creates an environment variable, myVar, in the environment of the script. The script, as it is currently presented, is functionally exactly equivalent to

#!/bin/bash

export myVar="myVal"

The fact that the export happens in the function body is not relevant to the scope of the environment variable (in this case). It will start to exist as soon as the function is called.

The variable will be available in the script's environment and in the environment of any other process started from the script after the function call.

The variable will not exist in the parent process' environment (the interactive shell that you run the script from), unless the script is sourced (with . or source) in which case the whole script will be executing in the interactive shell's environment (which is the purpose of "sourcing" a shell file).

Without the function call itself:

myFunc() {
    export myVar="myVal"
}

Sourcing this file would place myFunc in the environment of the calling shell. Calling the function would then create the environment variable.

See also the question What scopes can shell variables have?

You also have an alias in functons.shwith the same name as a function in your other file.

In functons.sh:

alias zzz=sz

In z.sh:

zzz () {
    df -h
}

This confuses bash.

Example:

$ cat f1
foo () { echo hello; }
alias xfoo=foo

$ cat f2
xfoo () { echo beep; }

$ source f1
$ source f2
$ shopt -s extdebug
$ declare -F foo
foo 1 f2

Without the xfoo alias in f1:

$ source f1
$ source f2
$ shopt -s extdebug
$ declare -F foo
foo 1 f1

The bash manual also includes the text

Aliases are confusing in some uses.

under the "BUGS" heading.