It is useful to consider some simple cases.
Consider for simplicity the case without clause (2), i.e. a formula like :
$\mathcal A := \lnot p_1$
and consider a formula $\mathcal B_1$.
Then :
$\mathcal A' := \lnot \mathcal B_1$.
We have an interpretation $I$ such that $I(p_1)=T$ and consider a new interpretation $I'$ such that $v_{I'}(\mathcal B_1)=T$.
You have to note that $\mathcal B_1$ is a formula built-up with connectives and some propositional letters, in general different from $p_1$; let $\mathcal B_1$ be $(q_1 \land ((\lnot q_2) \lor q_3))$.
$I'$ must be defined on $q_1,q_2,q_3$ and not necessarily on $p_1$; but this does not matter. The fact is that $I'$ "induces" (by the truth-table "calculations") a value for $\mathcal B_1$ and we know that this value is the same that $I$ assign to $p_1$, i.e. (in our example) $v_{I'}(\mathcal B_1)=I(p_1)=T$.
What we want to prove is that :
$v_{I}(\mathcal A)=v_{I'}(\mathcal A')$.
And this holds because we have :
$v_{I}(\mathcal A)=v_{I}(\lnot p_1)=F$
and :
$v_{I'}(\mathcal A')=v_{I'}(\lnot \mathcal B_1)=F$.
The next step is to "test the mechanism" on a slightly more complex formula like $\mathcal A := p_1 \land p_2$, where $\mathcal A' := \mathcal B_1 \land \mathcal B_2$.
Having completed this "warm-up", we have to prove it by induction on the "complexity" of $\mathcal A$, i.e. by induction on the number $n$ of occurrences of connectives in $\mathcal A$, where $\mathcal A$ is a formula built-up with propositional letters $\{ p_1, \ldots, p_k \}$.
The base case for $n=0$ is trivial; we have that $\mathcal A := p_1$ and $\mathcal A' := \mathcal B_1$.
The induction step must be performed for all the connectives of the language : $\lnot, \land, \lor$.
We have that $\mathcal A := \lnot \mathcal A_1$ or $\mathcal A := \mathcal A_1 \land \mathcal A_2$, and so on.
We assume as induction hypotheses that the property holds for the $\mathcal A_i$'s [which, of course, being subformulae of $\mathcal A$, have less occurrences of connectives than $\mathcal A$] and we use the "mechanism" above to conclude that it holds also for $\mathcal A$.
The successive step is to consider also the other propositional letters $q \notin \{ p_1, \ldots, p_k \}$.