Can $f$ be not computable even if $L$ is decidable?
If $L$ is decidable then you can use it's decider to compute $f$:
for the first symbol of $f(s)$ run the decider on the input $s\#x$ for any $x\in A$. For one of them, the decider will answer YES--this is the first letter in $f(s)$.

Continue doing this (e.g., for the second letter, decide if $s\#\#x \in L$, etc.), and reveal $f(s)$ letter-by-letter until the time where the decider answers NO for all $x\in A$, which means you reached the end of $f(s)$.

So if $L$ is decidable, $f$ is computable. Conversely, if $f$ is not computable, it can't be that $L$ is decidable.