... with //m modifier '.' doesn't match newline ...

Just to be clear: With or without the  //m regex modifier, the default behavior of the  . (dot) metacharacter is to match everything except a newline. It is only the  //s "dot matches all" modifier that causes dot to match absolutely everything.

my $d = $s =~ s/[\s\S]*^$m *: (.*)$(?:[\s\S]*)/$1/rm;

The expression  [\s\S] to express "match any character" cries out for comment. I assume it is used to avoid the  . (dot) metacharacter when promoted by  //s to "dot matches all" status.

This rubs me the wrong way. If dot (with //s) matches all, why not just use it that way? (All code examples that follow enable warnings and strictures. Also note that the  //r substitution modifier is only avaliable with Perl versions 5.14+.)

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s = qq{aaa       : AAA\n}
       . qq{bbb       : BBB\n}
       . qq{ccc       : CCC\n}
       ;
 print qq{[[$s]]};
 ;;
 my $m = 'bbb';
 ;;
 my $t = $s =~ s/.*^$m *: (.*)$(?:.*)/$1/rsm;
 ;;
 print qq{[[$t]]};
"
[[aaa       : AAA
bbb       : BBB
ccc       : CCC
]]
[[BBB
ccc       : CCC
]]
[download]

Consider the  (.*) capture group. With dot matching anything, it greedily grabs everything to the end of the string. To achieve an overall match, the regex still has to match  $ at the end of the string, which is easy, and  (?:.*) "zero or more of anything" after the end of the string, also easy. So capture group 1 and  $1 now contain everything to the end of the string, which is substituted back into the string.

But the intent of  (.*) was only to capture everything up to the  $ anchor before the first embedded newline (due to //m). How to restrain dot?

One way would be to use a  *? "lazy" modifier for the normally greedy  * match quantifier: dot will then match as little as necessary to get to the first  $ anchor.

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s = qq{aaa       : AAA\n}
       . qq{bbb       : BBB\n}
       . qq{ccc       : CCC\n}
       ;
 print qq{[[$s]]};
 ;;
 my $m = 'bbb';
 ;;
 my $t = $s =~ s/.*^$m *: (.*?)$(?:.*)/$1/rsm;
 ;;
 print qq{[[$t]]};
"
[[aaa       : AAA
bbb       : BBB
ccc       : CCC
]]
[[BBB]]
[download]

But one could argue that the intent of "anything except a newline" is more clearly expressed by  [^\n] and "capture as much as possible to the first newline" is better as  ([^\n]*) (remember that the code must be maintained, one must assume forever, so clear intent is important).

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s = qq{aaa       : AAA\n}
       . qq{bbb       : BBB\n}
       . qq{ccc       : CCC\n}
       ;
 print qq{[[$s]]};
 ;;
 my $m = 'bbb';
 ;;
 my $t = $s =~ s/.*^$m *: ([^\n]*)$(?:.*)/$1/rsm;
 ;;
 print qq{[[$t]]};
"
[[aaa       : AAA
bbb       : BBB
ccc       : CCC
]]
[[BBB]]
[download]

Lastly, an example in my own preferred style, taken from TheDamian's PBP:

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s = qq{aaa       : AAA\n}
       . qq{bbb       : BBB\n}
       . qq{ccc       : CCC\n}
       ;
 print qq{[[$s]]};
 ;;
 my $m = qr{ bbb }xms;
 ;;
 my $t = $s =~ s{ .* ^ $m [ ]* : [ ] ([^\n]*) $ .* }{$1}xmsr;
 ;;
 print qq{[[$t]]};
"
[[aaa       : AAA
bbb       : BBB
ccc       : CCC
]]
[[BBB]]
[download]

And that's the first several inches of the whole nine regex yards. HTH

But I see, that within square brackets the dot must be escaped.

Yes, you've got the idea its just the lingo you need help with now :) escaping means prefixing it with a backslash -- ie turn "." into "\." but that isn't required, as inside a character class [.] the dot is not a metacharacter, it is a literal character

(?s:.) means any character (including \n) alias [\w\W] alias [\s\S] alias [\d\D] alias \p{All}