It seems, I've mistaken. Here's my correction to my previous reasoning.

Let's present the re

'ab' =~ /((\w+)(?{print defined $2 ? "\$2=$2\n" : "\$2 not defined\n"})){2}/;

as

((\w+)(?{print...}))((\w+)(?{print...}))

Is \w{2} equivalent to \w\w, right? But we assume that the second copy of the re produces also the same $1 and $2 (not $3 and $4). Current position in the re marked with |.

1. First (\w+) captures all the text:
((\w+) | (?{print...}))((\w+)(?{print...}))
$2 receives the value 'ab', eval prints $2=ab.

2. Then we enter second copy of (\w+):
((\w+)(?{print...}))(( | \w+)(?{print...}))
$2 (and also $+, $^N, \2) receives the value undefined.

3. We see that \w not match. We do backtracking:
((\w+ | )(?{print...}))((\w+)(?{print...}))
We enter first copy of (\w+) from right to left, and $2 again receives the value undefined.

4. \w+ gives back the letter b (but $2 remains undefined, because we did not come left of the opening parenthesis for $2):
(( | \w+(?{print...}))((\w+)(?{print...}))
$2 remains undefined.

4. (\w+) captures none, because we did not come left of the opening parenthesis for $2:
((\w+) | (?{print...}))((\w+)(?{print...}))
$2 remains undefined. Eval prints $2=undefined.

5. Second copy of (\w+) captures the letter b:
((\w+)(?{print...}))((\w+) | (?{print...}))
Eval prints $2=b. Match successfull.