Proposition II.$3.2$ in Hartshorne
(1) This sort of argument is ubiquitous but thinking about it again I can understand the confusion. I don't know that I should do much more than rephrase: he needs to take in any affine open $\operatorname{Spec} A \subseteq X$ and show that $A$ is Noetherian. He shows that such a $\operatorname{Spec} A$ is locally Noetherian -- if he can show that this implies that $A$ is Noetherian then that certainly does the job.
You could view it merely as an attempt to cut down on notation, but this argument comes up a lot, and cutting down notation is usually good. Also, the fact that an open subscheme of a locally Noetherian scheme is locally Noetherian is a good one.
Vakil's book doesn't need more advertising but I think the way he codifies this sort of thing in Section 5.3 is worth seeing.
(2) $V$ is an open subset of $X$, so the structure sheaf should give you a restriction map \[ A= \mathscr{O}_X(X) \to \mathscr{O}_X(V) = B. \] Now a lot of implicit identifications enter and one has to become comfortable with them. The point is that $D(f)$ and $D(\bar{f})$ are the same open set. Remember: $D(f)$ is the set of points where $f$ doesn't vanish, and to check this at a particular point it doesn't matter if I restrict $f$ to some smaller neighborhood first.
Viewing the open set in these two different ways we see that the open subscheme is canonically isomorphic to both $\operatorname{Spec} A_f$ and $\operatorname{Spec} B_{\bar{f}}$, so the rings must be isomorphic. You could also see this as a special case of the fact that if $\pi\colon \operatorname{Spec} A \to \operatorname{Spec} B$ is a morphism coming from a map of rings $\phi\colon A \to B$ then $\pi^{-1}(D(f)) = D(\phi(f))$ for $f \in A$.
It might be worth writing this out while explicitly naming and using the isomorphisms of open subschemes with certain $\operatorname{Spec}$s, but you want to get away from that quickly.
The statement he is proving is:
If $X$ is locally Noetherian and $U = \mathop{\mathrm{Spec}}A\subseteq X$ is affine, then $A$ is a Noetherian ring.
He proves the following claims:
- If $X$ is locally Noetherian and $U = \mathop{\mathrm{Spec}}A\subseteq X$ is affine, then $U$ is locally Noetherian.
- If $U = \mathop{\mathrm{Spec}}A$ is affine and locally Noetherian, then $A$ is a Noetherian ring.
These claims imply the original statement. By "replacing $X$ by $U$", he means that in claim 2, he's calling the scheme $X$ instead of $U$. It may be confusing because it's $U$ that's being replaced by $X$ in claim 2, but he's probably thinking along the lines of reducing the original statement to the case that $X$ is affine (so $X$ is replaced by $U$, an affine scheme).
For your second question: as $X = \mathop{\mathrm{Spec}} A$ and $V = \mathop{\mathrm{Spec}}B$ are affine, the inclusion $i:V\hookrightarrow X$ induces a map of rings $i^\sharp:A\to B$. Define $\overline{f} = i^\sharp(f)$; this is the "image of $f$ in $B$".
Set $V_{\overline{f}} = \mathop{\mathrm{Spec}} B_{\overline{f}}$ and $X_f = \mathop{\mathrm{Spec}} A_f$, which we may identify with (principal) open sets in $X$. To show that $A_f\cong B_{\overline{f}}$, it suffices to show that $V_{\overline{f}} = X_f$. For $\mathfrak{p}\in V$, $\mathfrak{p}\in X_f$ if and only if $f\notin i(\mathfrak{p}) = (i^\sharp)^{-1}(\mathfrak{p})$, which happens if and only if $\overline{f} = i^\sharp(f)\notin \mathfrak{p}$, i.e., if and only if $\mathfrak{p}\in V_{\overline{f}}$. This shows the claimed equality because $X_f = D(f)\subseteq V$ by choice of $f$.
I would like to complement the answer to question 2 by expanding a little bit on Hoot's remark.
Let $X_f$ be the set of points of $X$ such that the germ of $f$ at each such point is not inside the maximal ideal of the corresponding local ring. If $U \cong \operatorname{Spec}B$ is an open affine set of $X$, then by exercise II.2.16(a) in Hartshorne, we have an isomorphism of schemes $X_f \cap U \cong D(\bar{f})$, where $\bar{f}$ is the image of $f$ under the ring homomorphism $A\rightarrow B$ that is associated to the inclusion $\operatorname{Spec}B \hookrightarrow \operatorname{Spec}A$.
Now, if it so happens that $X_f$ is a subscheme of $U$, then we have an isomorphism of schemes $X_f \cong D(\bar{f})$. If in addition $X=\operatorname{Spec}A$, then we get $\operatorname{Spec}A_f \cong \operatorname{Spec}B_{\bar{f}}$. Since this is an isomorphism of schemes, this gives $A_f \cong B_{\bar{f}}$.