Let $R$ be a ring, $S\subset R$ a multiplicative subset, and let $M$ be a Noetherian $S^{-1}R$-module, then does there exist some Noetherian $R$-module such that $S^{-1}N\cong M$? What about if we only consider localizations of the form $R_f$? What if we also require $R$ to be Noetherian?
If we let $N={_R M}$ then clearly $S^{-1}N\cong M$, but $N$ is very often not Noetherian.
You can always do this if $R$ is Noetherian. More precisely, if $M$ is a finitely generated $S^{-1}R$-module, you can always find a finitely generated $R$-module $N$ such that $S^{-1}N\cong M$. If $R$ is Noetherian, you can then conclude that $N$ is Noetherian.
To prove this, just take a finite set of generators for $M$ as an $S^{-1}R$-module and let $N$ be the $R$-submodule of $M$ they generate. Then $N$ is a finitely generated $R$-module, and the inclusion map $N\to M$ factors through the localization $N\to S^{-1}N$. The induced map $S^{-1}N\to M$ is surjective since $N$ generates $M$ as an $S^{-1}R$-module. It is injective since it is just the localization of the injection $N\to M$, and localization is exact. Thus $S^{-1}N\cong M$.
As an example to show that this can fail if $R$ is not Noetherian, let $R$ be any non-Noetherian domain and let $S^{-1}R$ be the field of fractions of $R$. Then $M=S^{-1}R$ is a Noetherian $S^{-1}R$-module. But if $S^{-1}N\cong M$, then any element of $N$ which is not annihilated by the localization $N\to S^{-1}N$ must generate a free submodule of $N$. Since $R$ is not Noetherian, this free submodule is not Noetherian, so $N$ is not Noetherian.
In particular, if $R$ is a valuation ring with value group $\mathbb{Q}$, then this gives an counterexample where $S$ is generated by a single element (since inverting any nonzero element of $R$ will give you the field of fractions).