Correct way of quoting command substitution - bash

Question

I have simple bash script which only outputs the filenames that are given to the script as positional arguments:
#!/usr/bin/env bash
for file; do
echo "$file"
done
Say I have files with spaces (say "f 1" and "f 2"). I can call the script with a wildcard and get the expected output:
$ ./script f*
> f 1
> f 2
But if I use command substitution it doesn't work:
$ ./script $(echo f*)
> f
> 1
> f
> 2
How can I get the quoting right when my command substition outputs multiple filenames with spaces?
Edit: What I ultimatively want is to pass filenames to a script (that is slightly more elaborate than just echoing their names) in a random order, e.g. something like that:
./script $(ls f* | shuf)

Answer 1

With GNU shuf and Bash 4.3+:
readarray -d '' files < <(shuf --zero-terminated --echo f*)
./script "${files[#]}"
where the --zero-terminated can handle any filenames, and readarray also uses the null byte as the delimiter.
With older Bash where readarray doesn't support the -d option:
while IFS= read -r -d '' f; do
files+=("$f")
done < <(shuf --zero-terminated --echo f*)
./script "${files[#]}"
In extreme cases with many files, this might run into command line length limitations; in that case,
shuf --zero-terminated --echo f*
could be replaced by
printf '%s\0' f* | shuf --zero-terminated
Hat tip to Socowi for pointing out --echo.

Answer 2

It's very difficult to get this completely correct. A simple attempt would be to use %q specifier to printf, but I believe that is a bashism. You still need to use eval, though. eg:
$ cat a.sh
#!/bin/sh
for x; do echo $((i++)): "$x"; done
$ ./a.sh *
0: a.sh
1: name
with
newlines
2: name with spaces
$ eval ./a.sh $(printf "%q " *)
0: a.sh
1: name
with
newlines
2: name with spaces

Answer 3

This feels like an XY Problem. Maybe you should explain the real problem, someone might have a much better solution.
Nonetheless, working with what you posted, I'd say read this page on why you shouldn't try to parse ls as it has relevant points; then I suggest an array.
lst=(f*)
./script "${lst[#]}"
This will still fail if you reparse it as the output of a subshell, though -
./script $( echo "${lst[#]}" ) # still fails same way
./script "$( echo "${lst[#]}" )" # *still* fails same way
Thinking about how we could make it work...

Answer 4

You can use xargs:
$ ls -l
total 4
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 1'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 2'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 3'
-rw-r--r-- 1 root root 0 2021-08-13 00:23 ' file 4'
-rwxr-xr-x 1 root root 35 2021-08-13 00:25 script
$ ./script *file*
file 1
file 2
file 3
file 4
$ ls *file* | shuf | xargs -d '\n' ./script
file 4
file 2
file 1
file 3
If your xargs does not support -d:
$ ls *file* | shuf | tr '\n' '\0' | xargs -0 ./script
file 3
file 1
file 4
file 2