C:\>perl -E "my $str1 = qq(|L|D|); my $str2 = qr(\|L\|);if ($str2 =~ $
+str1) { say 'foo';}"
 foo
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So why did the code I showed return true (or "foo")?"

Because the string defining the regex used in the  $str2 =~ $str1 expression (i.e., $str1) has the empty pattern as two (!) of its alternatives, and the empty pattern matches everything.

my $str1 = qq(|L|D|); has the empty pattern twice. It doesn't really matter what else is present. The stringization of the output of the  qr(\|L\|) expression is a bit complex, but it could be anything.

c:\@Work\Perl>perl -wMstrict -le
"my $str1 = qq(|L|D|);
 my $str2 = qr(\|L\|);
 print qq{stringization of qr// output: '$str2'};
 if ($str2 =~ $str1) { print 'foo'; }
"
stringization of qr// output: '(?^:\|L\|)'
foo

c:\@Work\Perl>perl -wMstrict -le
"my $str1 = qq(|X|Y|);
 my $str2 = qr(\|L\|);
 if ($str2 =~ $str1) { print 'foo'; }
"
foo

c:\@Work\Perl>perl -wMstrict -le
"my $str1 = qq(|X|Y|);
 my $str2 = qq(aaaaa);
 if ($str2 =~ $str1) { print 'foo'; }
"
foo
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Update: Changed  my $str2 = qq(xyzzy); in third code example above to  qq(aaaaa) to eliminate any question of case-insensitive matching.