General case:
We may assume that $P_{i}$ are finitely generated because $M$ is and $R$ is Noetherian.
From $\mathrm{Hom}(R^{n},R)\cong R^{n}$ we deduce that for finitely generated free module taking dual commute with any base change. Furthermore the same remains true for finitely generated projective modules as: $\mathrm{Hom}(-,R)$ and $-\otimes R'$ preserve split exact sequences, and there is always a natural morphism $\mathrm{Hom}_{R}(N,R)\otimes R'\to \mathrm{Hom}_{R'}(N\otimes R',R')$.
This being said let's consider a commutative diagram: $ \begin{matrix} P_{0}^{*}\otimes R' &\to & P_{1}^{*}\otimes R' & \to & D(M)\otimes R' & \to & 0\\ \downarrow \cong & & \downarrow \cong & & \downarrow & & \\ (P_{0}\otimes R')^{*} &\to & (P_{1}\otimes R')^{*} & \to & D(M\otimes R') & \to & 0\\ \end{matrix} $ where:
the upper line is exact by right-exactness of tensor product,
the bottom line is exact by the definition of $D(M\otimes R')$,
maps $(P_{i}\otimes R')^{*}\to P_{i}^{*}\otimes R'$ are natural isomorphisms from above,
rightmost vertical arrow is an induced map of cokernels.
Since the two indicated vertical arrows are isomorphisms so is the third (this can be viewed as a special case of five lemma, but also follows immediately from the universal property of cokernel).
The first attempt (uncomplete):
Warning: the proof below works only for flat ring extensions (what might be considered as a trivial case).
Since $R$ is noetherian $M$ is finitely presented, and so you can take $P_1, P_0$ to be finitely generated (hence likewise finitely presented). But for finitely presented modules taking dual commutes with flat base change. Therefore multiplying $0\to M^{*}\to P_{0}^{*}\to P_{1}^{*}\to D_{R}(M)\to 0$ by $R'$ we obtain an exact sequence $0\to (M\otimes R')^{*}\to (P_{0}\otimes R')^{*} \to (P_{1}\otimes R')^{*}\to D_{R}(M)\otimes R'\to 0$ where exactness on the left follows by right-exactness of the tensor product and left exactness of the dual.