Why does watch not show anything on screen when invoked with timeout in an script?

When I run this script, it does not display the watch output on the screen, and does not timeout after 5s.

§ cat script.sh
#!/bin/bash
timeout 5s watch -n 1 ps

This is what I see on the screen when I run the script:

However, running the content of the script directly in the terminal works as expected: i.e.,

§ timeout 5s watch -n 1 ps

What exactly is happening under the covers that is preventing watch from showing the content on the screen when invoked from a script?

Furthermore, using the --foreground option of timeout in the script makes it work as expected, but I don’t understand why.

If it matters, here are versions of the tools:

§ bash --version
bash --version
GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin19)
Copyright (C) 2007 Free Software Foundation, Inc.

§ watch -v
watch from procps-ng 3.3.12-dirty

§ timeout --version
timeout (GNU coreutils) 8.29
Copyright (C) 2017 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <https://gnu.org/licenses/gpl.html>.
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by Padraig Brady.

--foreground

    when not running timeout directly from a shell prompt, allow COMMAND to
    read from the TTY and get TTY signals; in this mode, children of
    COMMAND will not be timed out

$ script -c 'timeout --foreground 5s watch -n 1 ps'
$ less typescript