... what is the point of failure ...

There is no failure. As LanX pointed out, the regex is doing exactly what it's been told to do.

Win8 Strawberry 5.30.3.1 (64)  Sat 12/04/2021 21:26:27
C:\@Work\Perl\monks
>perl -Mstrict -Mwarnings
my $str='kklmn';

#        +--------------- if you look ahead to kkl
#        |
#        |  +---+-------- while also capturing it to capture group 1
#        |  |   |         (match point is not advanced)
#        |  |   |
#        |  |   |    +-------- and then if capture 1 captured anything
#        |  |   |    |
#        |  |   |    |  +----- match with what was captured to $1
#        |  |   |    |  |
#        |  |   |    |  |  +-- else match with mn
#        |  |   |    |  |  |
#        |  |   |    |  |  |    +------- and substitute this
#        |  |   |    |  |  |    |
#        v  v---v    v  v  v    v
$str=~ s/(?=(kkl))(?(1)\g1|mn)/***/g;

print "'$str' \n";
^Z
'***mn'
[download]

Update (many days later):

... what is the point of failure ...

That's a tautological mess.°

You've just invented a super complicated

$str =~ s/kkl/***/g;

Please try to explain what you expect in an SSCCE

°) actually it's legal code, a conditional regex which is only true after the group inside lookahead assertion matched, but can't ever be false because the former assertion skips those positions already 🤷🏻‍♂️

Note that this was also posted in p5p, where it was noted that the example as given does not compile.

> the example as given does not compile.

there is a quote missing, yes!

... apart from this it "works".

  DB<324> $str='kklmn'  # end quote added

  DB<325> $str=~ s/(?=(kkl))(?(1)\g1|mn)/***/g

  DB<326> print $str
***mn
  DB<327>
[download]

Cheers Rolf
_{(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery}