Let S be a multiplicative set in a ring R. We can see that every finitely generated $(S^{-1})R$-module is a localization of a finitely generated R-module.
Then, more generally, is every finitely n-presented $ (S^{-1})R$-module a localization of a finitely n-presented R-module ?
No. Let $R = k[x, y_i, z_i]/(xy_i, y_iz_i)$ where $i = 1, 2, 3, \ldots$ Let $S$ be the multiplicative system regenerated by $z_1, z_2, z_3, \ldots$ Then $x$ becomes a nonzero divisor in the ring $R' = S^{-1}R$. Hence the module $M' = R'/xR'$ is pseudo-coherent, i.e., it is finitely $n$-presented for every n.
But there does not exist a finitely 3-presented module $M$ over $R$ such that $M'$ is the localization of $M$. First, note that there does exists a finitely presented module, namely $R/xR$, whose localization is $M'$. Next, let $R_n$ be the ring where we invert $z_1, z_2, \ldots, z_n$ in $R$. Denote $M_n = M \otimes R_n$. Because $R' = \text{colim} R_n$ for large enough $n$ we have $M_n \cong R_n/xR_n$ for example by Lemma Tag 05N7 of the Stacks project. Since $R \to R_n$ is flat, this would imply that $R_n/xR_n$ is finitely $3$-presented. But it isn't, because the kernel of $R_n \stackrel{x}{\to} R_n$ is not finitely generated.
Remark 1: It is often possible to construct counter examples to (false) general commutative algebra statements by the procedure above.
Remark 2: A true statement is that if one has an $(-n)$-pseudo coherent module $M'$ over $R'$ and $R'$ is the filtered colimit of rings $R_i$, then there exists an $i$ and a finite complex of finite free modules $E$ over $R_i$ whose derived base change $E' = E \otimes^L_{R_i} R'$ has cohomology $H^0(E') = M'$ and vanishing $H^p(E')$ for $-n < p < 0$ and $p > 0$.
Remark 3: In the localization situation above one can get the complex $E$ of the previous remark to live over the ring $R$ by getting rid of denominators in the obvious way.
