Another way, using a positive look-ahead assertion and the  \K special escape of 5.10+ (see  (?<=pattern) \K in Look Around Assertions) which is effectively a variable-width positive look-behind (an actual look-behind in Perl regex may only be fixed-width):

>perl -wMstrict -le
"my $s =
   'there is no FOO stuff i do not want BAR unwanted BAR stuff here';
 ;;
 my $t = $s;
 $t =~ s{ FOO \K .*? (?= BAR) }{}xms;
 print qq{'$t'};
 ;;
 $t = $s;
 $t =~ s{ FOO \K .* (?= BAR) }{}xms;
 print qq{'$t'};
"
'there is no FOOBAR unwanted BAR stuff here'
'there is no FOOBAR stuff here'
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The given examples illustrate the effect of the  ? quantifier modifier to achieve 'lazy' matching. If the substitution is to be done repeatedly, the  /g regex modifier should be used.

As suggested above, a positive look-behind could have been used in the examples because the pattern in question is fixed-width, but look-behind assertions, positive or negative, cannot be used for general patterns, which seemed to be the thrust of the OP.

If I understand your question correctly:

$string=~s/(pattern1).*(pattern2)/$1$2/s;
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I need a regex, a regex is what I need
Well I need a regex, a regex is is what I need
And I said I need regex, regex; a regex is what I need
And if I share with you my story would you share your regex with me

-- Aloe Blacc - I need a dollar

Like said above, it sounds like you want the /s modifier.

use strict;
use warnings;

my $foo = "blabla foo \n yadayada \n bar bla bla";
print "\n<--\n$foo\n-->\n";

$foo =~ s/foo.+bar//s;
print "\n<--\n$foo\n-->\n";
__END__
Output:

<--
blabla foo 
 yadayada 
 bar bla bla
-->

<--
blabla  bla bla
-->
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