I'm not sure how you add on paper, but for me, I only ever consider one digit of the number at a time. From my perspective, the problem is that your inputs can be arbitrary numbers, not just digits.

To get all elements to be digits, you can:

• work in a base greater than the largest element (changes the rules, feels like cheating)
• preprocess the arrays
• treat the elements with more care when you pop them

Here's an example of the last approach:

#!/usr/local/bin/perl
use strict;
use warnings;

use constant BASE => 10;

run( @$_ ) for (
    [    'case1 when @a > @b',
        [qw(1 0 9 15 6)],
        [qw(    1 1  4 )], ],
    [   'case2 when @a < @b',
        [qw(   1  5 9)],
        [qw(10 9  3 6)], ],
    [    'case3 when @a == @b',
        [qw(1 1 6 9 2)],
        [qw(2 1 4 4 8)], ] );


sub run {
    my ($msg, $a, $b) = @_;

    local $\ = "\n"; # or use say on 5.10

    print "# $msg";

    print "@$a";
    print "@$b";
    print "__________+";

    my @result = add( $a, $b );
    print "@result";
    print "";
}

{
    use integer 'division';
    
    sub add {
        my ($a, $b) = @_;
        my @sum;
        my $carry = 0;

        while (@$a || @$b || $carry) {
            my $_a = rem_lsd($a);
            my $_b = rem_lsd($b);

            my $sum = $_a + $_b + $carry;

            $carry = $sum / BASE;
            $sum %= BASE;

            unshift @sum, $sum;
        }

        return @sum;
    }
    
    # get (remove) least significant digit
    # corrects for multidigit inputs (by removing the lsd from the las
+t element)
    # returns zero when array is empty
    sub rem_lsd {
        my $elem = pop @{$_[0]} || 0;

        # put back extra digits
        if (my $extra = $elem / BASE) {
            push @{$_[0]}, $extra;
        }
        $elem %= BASE;

        return $elem;
    }
}
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