s Treat string as single line. That is, change "." to match any character whatsoever, even a newline, which normally it would not match.

You're not taking into account the non-greediness. To accommodate matching the $ (which is before the newline) $1 holds 'foo'. If you also want to match the terminal newline, use \z instead of $:

$ perl -e '$_="foo\n"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo|

$ perl -e '$_="foo\n"; print "1=|$1|\n" if m/(fo.+?)\z/s'
1=|foo
|
[download]

Your comparison with the first one-liner is not comparing apples with apples:

$ perl -e '$_="foo\nbar"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo
bar|

$ perl -e '$_="foo\nbar\n"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo
bar|

$ perl -e '$_="foo\nbar\n"; print "1=|$1|\n" if m/(fo.+?)\z/s'
1=|foo
bar
|
[download]

I also second ++Fletch's recommendation to use Regexp::Debugger. This allows you to step through the matching process and see exactly what's happening. I often use it myself.

— Ken

I get what's going on now, thank you.

I still think it's a bug, or at the very least a major implementation quirk that is incompatible with other regex implementations. Javascript and Golang treat /s the intuitive way and don't make an exception for \n at the end of a string.

Type this into any browser console:

"foo\nbar".match(/(fo.+?)$/s) && RegExp.$1
"foo\n".match(/(fo.+?)$/s) && RegExp.$1
[download]

Or try it on regex101.com - you get different matching results on PCRE vs non-PCRE based engines.

I think it warrants a big warning in perlre. It was a nasty surprise for me even though I am far from being a perl novice.