Let $f\colon\operatorname{Spec}A\to\operatorname{Spec}B$ be a morphism of affine schemes, corresponding to the ring homomorphism $\varphi\colon B\to A$ and let $M$ be an $A$-module. I'm trying to understand why $$f_*\widetilde{M}\cong\widetilde{M_{/B}},$$ where the index $_{/B}$ denotes restriction of coefficients.
Let $g$ be an element of $B$. Then $\widetilde{M_{/B}}(D(g))=(M_{/B})_g$ by definition. On the other hand, $f_*\widetilde{M}(D(g))=\widetilde{M}(D(\varphi(g))=M_{\varphi(g)}$. So what I need to show is that $M_{\varphi(g)}\cong(M_{/B})_g$ as $B_g$-modules.
We have a morphism $\varphi:B\to A$, and using this $M$ has a natural structure as a $B$-module, which we call $M_{/B}$, where the action of $B$ on $M_{/B}$ is given by $b\cdot m=\varphi(b)m$. Note that $M_{/B}$ is still the same set as $M$, we are just emphasizing a different module structure with this notation. So if you want to localize $M_{/B}$ at $g$, then really you're just localizing $M$ at $\varphi(g)$.
For instance, if you want to act on $M_{\varphi(g)}$ by an element of $B_g$, how do you do it? You take $$\frac b{g^n}\cdot\frac m{\varphi(g)^k}=\frac{b\cdot m}{g^n\cdot\varphi(g)^k}=\frac{\varphi(b)m}{\varphi(g^n)\varphi(g)^k}=\frac{\varphi(b)m}{\varphi(g)^{n+k}}.$$
So clearly there's no difference in module structure, the only difference is writing $g$ or $\varphi(g)$ in the quotients.
If you want to be precise, you can define a map $M_{\varphi(g)}\to (M_{/B})_g$ by sending $\frac m{\varphi(g)^n}$ to $\frac{m}{g^n}$. It's definitely a bijection of sets, so you just need to check it's $B_g$-linear.