I think my trouble is in understanding what $a and $b are referring to.

They are referring to the elements you pass into the sort routine. Which element of that list gets associated when to $a or to $b is sort's business. doc;//sort assigns elements of that list to $a and $b as aliases, as the underlying sort algorithm demands.

You were passing keys %HoH to the sort function, hence the keys of the outer hash, the chars (a .. g) in the order perl's hashing implementation sees fit. $a and $b, inside the sort function block, will never be anything else than an alias to an element of the list you passed in.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

I find it hard to comprehend how sort can "see outside" of the foreach loop.

That's common for all blocks.

my $var = "Hi!";
{
   print("$var\n");
}
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my $var = "Hi!";
for (1..2) {
   print("$var\n");
}
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my $var = "Hi!";
print(map { "$var\n" } 1..2);
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my $var = "Hi!";
sub func {
   print("$var\n");
}
func();
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Yes, the following works even though $var goes out of scope before func is called.

{
   my $var = "Hi!";
   sub func {
      print("$var\n");
   }
}

func();
[download]