Would this proof work?
Given a language $L$ that is Turing recognizable and a TM M that recognizes it and a homomorphism $f$, we build a NTM M' that recognizes $f(L)$,
M' looks like this:
On input $w$ :
-Non-deterministicly feed words from $\Sigma^{*}$ to $f$ until you obtain w.
-run M on $f^{-1}(w)$, if M accepts accept, otherwise reject.
Your proof has a flaw.
Assume $L=\{b\}$, and $f(a)=f(b)=w$ for some word $w$. We have $f(L)=\{w\}$.
TM $M'$ on $w$ finds the preimage $a$, then checks $a\not\in L$, and rejects. This is wrong, since $w$ should be accepted.
The point is, we should not reject $w$ just because we found a preimage of $w$ which is not in $L$. We should continue checking all such preimages.
Above, doing that we would consider $b$ as well, and cause $w$ to be accepted.
The problem with your proof is where you say "until you obtain $w$". That makes it sound to me like you stop the search as soon as you find a single $x$ such that $f(x)=w$. If that is what your machine $M'$ does, then your proof is faulty, for the reasons chi explains.
Also, there is an additional step that is worth explaining: it's worth mentioning that any language that's recognizable by a NTM is also recognizable by a standard deterministic TM. This is a standard fact that probably doesn't need to be proven here.
Fortunately, there is a simple fix. Here is an improved proof:
Given a language $L$ that is Turing recognizable and a TM $M$ that recognizes it and a homomorphism $f$, we build a NTM $M'$ that recognizes $f(L)$.
$M'$ looks like this:
On input $w$ :
Non-deterministicly guess a word $x \in \Sigma^*$.
If $f(x)=w$ and $M$ accepts on input $x$, accept, otherwise reject.
This works because $M'$ is a non-deterministic Turing machine. A NTM accepts $w$ if there is at least one branch accepting $w$, so if there is any word $x$ such that $f(x)=w$ and $M(x)$ accepts, this will find it and accept.
Moreover, any language that can be recognized by a nondeterministic TM, can be recognized by a deterministic TM. It follows that $f(L)$ is Turing recognizable.
Credit: thanks to xskxzr for explaining the idea of the proof and for the improved formatting.
