As I understand your task (which seems to be different from BrowserUk), this should do it:

#!/usr/bin/perl -w
use strict;

my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4);

my $input = "abcdaeec";

for (2, 3) {
    print "n=$_: ";
    my $n = $_ - 1;

    my @sums;
    while ( $input =~ /(.)(?=.{$n}(.))/g ) {
        push @sums, $hash{$1} + $hash{$2};
    }
    print "@sums\n";
}

__END__
n=2: 7 4 7 7 6 9
n=3: 5 3 9 7 7
[download]

Update: let YAPE::Regex::Explain explain what the regex does:

use YAPE::Regex::Explain;

print YAPE::Regex::Explain->new( qr/(.)(?=.{1}(.))/ )->explain;

__END__
The regular expression:

(?-imsx:(.)(?=.{1}(.)))

matches as follows:
  
NODE                     EXPLANATION
----------------------------------------------------------------------
(?-imsx:                 group, but do not capture (case-sensitive)
                         (with ^ and $ matching normally) (with . not
                         matching \n) (matching whitespace and #
                         normally):
----------------------------------------------------------------------
  (                        group and capture to \1:
----------------------------------------------------------------------
    .                        any character except \n
----------------------------------------------------------------------
  )                        end of \1
----------------------------------------------------------------------
  (?=                      look ahead to see if there is:
----------------------------------------------------------------------
    .{1}                     any character except \n (1 times)
----------------------------------------------------------------------
    (                        group and capture to \2:
----------------------------------------------------------------------
      .                        any character except \n
----------------------------------------------------------------------
    )                        end of \2
----------------------------------------------------------------------
  )                        end of look-ahead
----------------------------------------------------------------------
)                        end of grouping
----------------------------------------------------------------------
[download]

And see perlop for what m//g does in scalar context (like the while (...) here).

$s = "abcdef"; 
print "$1,$2\n" while $s =~ /(.)(?=.(.))/g;
[download]

"Sliding window" is a decptive title, but the problem is interesting.

A not very perlish solution:

#! perl -slw
use strict;

my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4);

my $input = "abcdaeec";
our $N //= 2;

$,= ' ';
print map{
    my $iplus = ord() - ord('a');
    $iplus = ( $iplus + $N ) % keys %hash;
    $iplus = chr( ord('a') + $iplus );

    $hash{ $_ } + $hash{ $iplus };
} split '', $input;

__END__
C:\test>junk41 -N=2
7 4 9 5 7 5 5 9

C:\test>junk41 -N=3
5 5 7 4 5 9 9 7

C:\test>junk41 -N=4
6 3 6 8 6 7 7 6
[download]

my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4);
my $input = "abcdaeec";
my @input = split //, $input;

for my $w (2, 3) {
    for (my $i = $w; $i < @input; $i++) {
        printf "%s+%s = %d+%d = %d\n",
               $input[$i-$w], $input[$i],
               $hash{$input[$i-$w]}, $hash{$input[$i]},
               $hash{$input[$i-$w]} + $hash{$input[$i]};
    }
}

__END__
a+c = 2+5 = 7
b+d = 1+3 = 4
c+a = 5+2 = 7
d+e = 3+4 = 7
a+e = 2+4 = 6
e+c = 4+5 = 9
a+d = 2+3 = 5
b+a = 1+2 = 3
c+e = 5+4 = 9
d+e = 3+4 = 7
a+c = 2+5 = 7
[download]

my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4);
my $input = "abcdaeec";

my @v = map $hash{$_}, (split //, $input);

for my $n (2, 3)
{
    my @w = map $v[$_] + $v[$_ + $n], 0 .. $#v - $n;
    print "@w\n";
}

__END__
7 4 7 7 6 9
5 3 9 7 7
[download]