Definition of Zariski localization along a closed subset

Let me answer question 1. The short answer is: "this is essentially by definition of locally prinicipal, and the facts that ${\rm Spec}A$ is quasi-compact, and principal opens form a basis for its topology".

The extended version of an answer is: Let $\mathscr{I}$ denote the ideal sheaf on ${\rm Spec} A$ corresponding to $I$. The fact that $I$ is locally principal means precisely that that locally on ${\rm Spec} A$, $\mathscr{I}$ is generated by one global section. THus there is an open covering ${\rm Spec} A = \bigcup_{i \in I} U_i$ such that $\mathscr{I}|_{U_i}$ is generated -- as an $\mathcal{O}_{U_i}$-module -- by one global section. Note that this property remains true after restricting to an open subset of $U_i$. Thus, and as the principal open subsets form a basis for topology of ${\rm Spec} A$, we can refine our covering to one by principal opens $D(g_i) \cong {\rm Spec}A[g_i^{-1}] \subseteq {\rm Spec} A$. Moreover, as ${\rm Spec} A$ is quasi-compact, we can pick a finite subcovering and so we have finitely many $g_i$'s such that ${\rm Spec} A = \bigcup_i D(g_i)$, which is equivalent to the claim that the unit ideal of $A$ is generated by the $g_i$'s. Now $\mathscr{I}|_{D(g_i)}$ is the ideal sheaf on $D(g_i) = {\rm Spec}A[g_i^{-1}]$ corresponding to the $A[g_i^{-1}]$-module $IA[g_i^{-1}]$, and and it is generated by one global section by construction. This implies that $IA[g_i^{-1}] = \Gamma(D(g_i), \mathscr{I})$ is generated (as $A[g_i^{-1}]$-module) by one element.

Regarding question 2:

My interpretation of the phrase (which fits the proof, i.e. makes the proof of the Lemma work) is the following:

Given a ring $B$ and an ideal $I \subseteq B$, the localization of $B$ along $V(I)$ is $S^{-1}B$ where $S=B \setminus \bigcup V(I)=B \setminus \bigcup_{\mathfrak{p}\in V(I)} \mathfrak{p}$ (note that $S$ is multiplicatively closed).

This is, firstly, ind-Zariski localization of the ring $B$ as per the mentioned Definition 2.2.1.

Secondly, it does what is claimed in the proof of Lemma 3.1, namely $IS^{-1}B$ is contained in the radical:

Indeed, primes of $S^{-1}B$ correspond to primes $\mathfrak{p}$ of $B$ that are contained in $\bigcup V(I)$. Given such a prime, the inclusion $\mathfrak{p}+I \subseteq \bigcup V(I)(=\bigcup_{I \subseteq \mathfrak{q} \in \mathrm{Spec}\,B}\mathfrak{q})$ shows that $\mathfrak{p}+I$ does not contain $1$, so there is a maximal ideal $\mathfrak{m}$ of $B$ with $\mathfrak{p}, I \subseteq \mathfrak{m}$. This shows that every prime ideal of $S^{-1}B$ is contained in a maximal ideal coming from $V(I)$, so in particular, every maximal ideal of $S^{-1}B$ contains $I$.

In fact, this argument shows that $\mathrm{Spec}\,S^{-1}B=\bigcup_{\mathfrak{p}\in V(I)}\mathrm{Spec}\,A_{\mathfrak{p}}$ as you mention.