One-way functions and P=NP This site contains various discussions of one-way functions and their relation to P versus NP.

Some of these discussions use a language 
$L=\{(x',y) ~\mid~ x'\le x \text{ and } f(x)=y \}$, where $f:\Sigma^*\to\Sigma^*$ is the one-way function and $x'\le x$ is the prefix relation.
Now one central claim is that this language $L$ is contained in NP, since the word $x$ is a YES-certificate for $(x',y)\in L$.

I do not see why this claim is justified.
Why is the length of the certificate $x$ polynomially bounded in the length of $(x',y)$?  

Couldn't it be possible that $x$ is exponentially long in $y$ and $x'$, but $f(x)$ is short and quickly computable from $x$?
Yes, it could be that in the language you give, $x$ is exponentially long in $(y,x')$, and $f$ is an efficiently computable one-way function (note that it only has to run in time polynomial in its input length, so $f(x)$ needs not be computable in time polynomial in $(y,x')$).

However, this is really a minor issue: the answers to this question that you read are simply a bit informal, and only give an intuition of the proof that OWF implies $P \neq NP$. Intuitively, to fix this, modify your language as follows:

$L=\{(1^n, x',y) ~\mid~ \exists x, |x| = n, x'\le x, \text{ and } f(x)=y \}$, 

where $1^n$ means a sequence of $n$ consecutive one, which exactly allows to fix the issue you point out (note that here $x'\le x$ means $x'$ is a prefix of $x$).

Note: the second answer to the question you link to does provide a link to an exercise sheet which contains the more formal solution.