Maybe I just don't understand the question, but assuming  unpack '%32b*', ...   is sufficiently efficient, wouldn't a count on the  'b' template specifier serve to count the set bits from the beginning of the string up to but not including a given bit offset?

>perl -wMstrict -le
"my $b = qq{\x01\x03\x07\xff\xfe\xfc};
 ;;
 print '          1111111111222222222233333333334444444444';
 print '0123456789' x 5;
 print unpack 'b*', $b;
 print ''
 ;;
 for my $offset (0, 1, 7 .. 10, 15 .. 20, 24 .. 28, 30 .. 32) {
   my $c = unpack qq{%32b$offset}, $b;
   my $vc = vec_sum($b, $offset);
   die qq{pack sum $c != vec sum $vc} if $c != $vc;
   printf qq{before bit at offset %2d: %2d bits set \n}, $offset, $c;
   }
 ;;
 sub vec_sum {
   my ($v, $p) = @_;
   ;;
   my $c = 0;
   vec($v, $_, 1) and ++$c for 0 .. $p - 1;
   return $c;
   }
"
          1111111111222222222233333333334444444444
01234567890123456789012345678901234567890123456789
100000001100000011100000111111110111111100111111

before bit at offset  0:  0 bits set
before bit at offset  1:  1 bits set
before bit at offset  7:  1 bits set
before bit at offset  8:  1 bits set
before bit at offset  9:  2 bits set
before bit at offset 10:  3 bits set
before bit at offset 15:  3 bits set
before bit at offset 16:  3 bits set
before bit at offset 17:  4 bits set
before bit at offset 18:  5 bits set
before bit at offset 19:  6 bits set
before bit at offset 20:  6 bits set
before bit at offset 24:  6 bits set
before bit at offset 25:  7 bits set
before bit at offset 26:  8 bits set
before bit at offset 27:  9 bits set
before bit at offset 28: 10 bits set
before bit at offset 30: 12 bits set
before bit at offset 31: 13 bits set
before bit at offset 32: 14 bits set
[download]

Update: Example code changed to include check against  vec sum algorithm given in OP; also made things a bit prettier.