No. Let R = k[x, y_i, z_i]/(xy_i, y_iz_i) where i = 1, 2, 3, ... Let S be the multiplicative system regenerated by z_1, z_2, z_3, ... 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, ..., z_n in R. Denote M_n = M &otimes; R_n. Because R' = colim R_n for large enough n we have M_n &cong; R_n/xR_n for example by <a href="http://stacks.math.columbia.edu/tag/05N7">Lemma Tag 05N7</a> of the Stacks project. Since R ---> 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 --- x ---> 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;<sup>L</sup><sub>R_i</sub> 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.