I don't think there is any direct way of doing this with a regex as there is no way to do a capture of the encompassed string without advancing the position.

Thus, you have to reset the position (pos) to the end of the first bracketing word ($+[1]) after each match.

$s = 'The quick brown fox jumped over the other quick brown fox';

print $2 and pos($s) = $+[1] 
    while $s =~ m[\b(\S+)\b(.+?)\b\1\b]g;

 brown fox jumped over the other
 fox jumped over the other quick
 jumped over the other quick brown
[download]

this includes the white space either end of the encompassed text. To avoid that change the regex to  m[\b(\S+)\b\s+(.+?)\s+\b\1\b]g.

You want capture and lookahead, print $2, $/ while /(\b\w+\b)(?=(.*?)\1)/g; \1 there is the form of backreference needed in regexen.

Very cool, but the output of print $2, $/ while /(\b\w+\b)(?=(.*?)\1)/g; is:

 brown fox jumped over the other
 fox jumped over the other quick
 jumped over the other quick brown
 o  #????
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In order to pass over the "o" (from "the o_the_r") the backreference in the lookahead needs to be anchored to word boundaries, ie:

print $2, $/ while /(\b\w+\b)(?=(.*?)\b\1\b)/g;

Not to mention, what is the desired output if the string is "the one the two the"?

I can't believe you missed out on this opportunity to wave (--) the magic wand variable ($|) to get only the odd or even elements of a list:

print grep --$|,($|||=1)&& /\b(\w+)\b(?=(.*?)\b\1\b)/g'
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What the good lord does that code do? Or rather, how does it work? Whats the $|||=1 bit, etc?