Match a word if it follows itself:
/\b(\w+)\W\1\b/
Correct the problem (remove the second word):
s/\b(\w+)\W\1\b/$1/g

\b matches a word break. It is zero bytes long and lifes on both sides of a word. In the words "|sides| |of| |a| |word|", each | shows the position of a \b.

The regex matches a word break (which is also before the first and behind the last word in a string/on a line), followed by a word (\w+) containing one or more of [a-zA-Z0-9_] followed by a non-word-char (\W) (which is everything but the chars listed before).

You know $1, which contains the contents of the first ( ) in the last executed regex? \1 is the same, but it's for the current regex and contains the word we found before by using \w+.

Adding another word-break at the end keeps us from matching /in/information/, because (\w+)\W\1 would match the in (as \w+ in ( ) going to \1), the / as match for \W and "in" of "information" for \1, because \1 contains the "in" from the \w+ match.

The first regex could be used on your whole text, there is no need to split it into lines.
The second regex removes any duplicate words converting /help/one/one/one/something to /help/one/something

Speaking of texts, some suggestions:

you may want to add a "+" after \W which would also match "text, text"

replace the \W by \/ to match only / instead of every non-word char

replace \w by [^\/] to allow everything between two / which is no /, but remember to also replace \b by \/ because \b won't work any longer