Hartshorne Lemma II.5.3 Proof The question I have is from the proof of the lemma above, but it is actually a more general statement about quasi-coherent sheaves on an affine scheme. Suppose $X= \text{Spec }A$ for some ring $A$, and $\mathscr{F}$ is a quasi-coherent sheaf on $X$. Then for some open affine cover of $X$, the restriction sheaf is isomorphic to a sheaf of a module over the corresponding ring. In particular, if $\text{Spec }B$ is in the cover, then $\mathscr{F}|_{\text{Spec } B} \cong \widetilde{M}$ for a $B$-module $M$. This part is by definition.
Now $\text{Spec }B$ is covered by distinguished open sets of the form $D(g)$ for $g\in A$, and for any such open set the inclusion $D(g)\subseteq \text{Spec }B$ is induced by the ring map $B\to A_g$. Thus $\mathscr{F}|_{D(g)} \cong (M\otimes_B A_g)^{\tilde{}}$. 
He deduces the last sentence from a previous proposition that deals with properties of the sheaves of modules. The two properties that seem important for this deduction are the following: For a ring map $A \to B$ inducing the map of spectra $f:\text{Spec }B \to \text{Spec }A$, 
(1) If $M$ and $N$ are $A$-modules, then $(M\otimes N)^{\tilde{}} \cong \widetilde{M} \otimes_{\mathcal{O}_{\text{Spec }A}} \widetilde{N}$.
(2) For any $A$-module $M$, $f^*(\widetilde{M})\cong (M\otimes_{A} B)^{\tilde{}}$.
I cannot seem to make the connection. So any help with his last statement would be great. Thanks.
 A: I am clarifying further the crucial point raised in the comments to @hwong557's answer.
Let $(f,f^\#): (X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$ be a morphism of locally ringed spaces. Suppose that $f(X)$ is open in $Y$ and that $f$ induces an isomorphism $\big(X,\mathcal{O}_X\big) \rightarrow \big(f(X),\mathcal{O}_Y|_{f(X)}\big)$ of locally ringed spaces. Hence, we have an isomorphism of sheaves $\mathcal{O}_Y|_{f(X)} \cong f_*\mathcal{O}_X$. Moreover, for every open set $U$ of $X$ we have an isomorphism of rings $\mathcal{O}_Y(f(U)) = \mathcal{O}_Y|_{f(X)}(f(U)) \cong  f_*\mathcal{O}_X(f(U)) = \mathcal{O}_X(U)$.
Now let $\mathscr{G}$ be an $\mathcal{O}_Y$-module. Let us look at the sections of the presheaf that gives rise to $f^* \mathscr{G}$. Let $U$ be open in $X$. Then the sections of that presheaf over $U$ are $f^{-1}\mathscr{G}(U) \otimes_{f^{-1} \mathcal{O}_Y(U)} \mathcal{O}_X(U)$. But $f^{-1} \mathcal{O}_Y(U) = \lim_{V' \supset f(U)}  \mathcal{O}_Y(V') =\mathcal{O}_Y(f(U))$, and we already know that this latter ring is isomorphic to $\mathcal{O}_X(U)$. Hence, $f^{-1}\mathscr{G}(U) \otimes_{f^{-1} \mathcal{O}_Y(U)} \mathcal{O}_X(U) \cong f^{-1}\mathscr{G}(U)$. This now shows that $f^* \mathscr{G}$ is the same as $f^{-1}\mathscr{G}$, and the latter is by definition $\mathscr{G}|_X$.
A: Let $\phi: Spec (A_g) \to Spec (B)$ be the inclusion map. Then:
$$\mathcal F\mid_{D(g)} =
(\mathcal F\mid_{V})\mid_{D(g)} = 
 (\tilde M) \mid_{D(g)} = \phi^*(\tilde M) = (M \otimes_B A_g)^\tilde{}
$$
