It would have been better indeed if Jech had inserted an explanatory paragraph or referred the reader to textbooks on model theory to look up the details of the proof.
Apart from the usual method of structural induction on the complexity of $\varphi$, it should be remarked that the lemma has an argument form that is interesting in itself.
The lemma can be expressed in the language of propositional calculus as
$$P\leftrightarrow (Q\rightarrow R)$$
We can break $P$ down further, considering that $\mathcal{B}$, a submodel of $\mathcal{A}$, is called elementary, denoted by $\mathcal{B}\preccurlyeq\mathcal{A}$, if and only if, given any elements $a_1, \ldots, a_n \in B$ and any formula $\varphi$,
$$\mathcal{B}\models\varphi(a_1,\ldots,a_n)\iff\mathcal{A}\models \varphi(a_1,\ldots,a_n)$$
Thus, we can rewrite the lemma replacing $\mathcal{B}\preccurlyeq\mathcal{A}$ with the aforementioned equivalence
$$(P_{B}\leftrightarrow P_{A})\leftrightarrow (Q\rightarrow R)$$
As noted in the question, given a subset $\mathrm{B}\subseteq\mathrm{A}$ and any formula $\varphi(x, a_1,\ldots, a_n)$ where $a_1,\ldots,a_n\in\mathrm{B}$, it is relatively straightforward to show that
$$(P_{B}\leftrightarrow P_{A})\rightarrow (Q\rightarrow R)$$
We shall work out
$$(Q\rightarrow R)\rightarrow(P_{B}\leftrightarrow P_{A})$$
that is, assuming the implication of the lemma.
In case that $\varphi$ is an atomic (quantifier-free and connective-free) formula,
$$\mathcal{B}\models\varphi(a_1,\ldots,a_n)\implies\mathcal{A}\models \varphi(a_1,\ldots,a_n)$$
is agreeably obvious by the given conditions. We get
$$\mathcal{A}\models\varphi(a_1,\ldots,a_n)\implies\mathcal{B}\models \varphi(a_1,\ldots,a_n)$$
by using the assumed implication of the lemma. Hence,
$$\mathcal{A}\models\varphi(a_1,\ldots,a_n)\iff\mathcal{B}\models \varphi(a_1,\ldots,a_n)$$
In case that $\varphi$ is of the form $\neg\psi$, we observe
$$\mathcal{B}\models\neg\psi(a_1,\ldots,a_n)\iff\mathcal{B}\nvDash\psi(a_1,\ldots,a_n)\iff\mathcal{A}\nvDash\psi(a_1,\ldots,a_n)\iff\mathcal{A}\models\neg\psi(a_1,\ldots,a_n)$$
In case that $\varphi$ is of the form $\psi\wedge\theta$
$\mathcal{B}\models\psi(a_1,\ldots,a_n)\wedge\theta(a_1,\ldots,a_n)\iff\mathcal{B}\models\psi(a_1,\ldots,a_n)\text{ and } \mathcal{B}\models\theta(a_1,\ldots,a_n)$
$\iff\mathcal{A}\models\psi(a_1,\ldots,a_n)\text{ and }\mathcal{A}\models\theta(a_1,\ldots,a_n)$
$\iff\mathcal{A}\models\psi(a_1,\ldots,a_n)\wedge\theta(a_1,\ldots,a_n)$
In case that $\varphi$ is of the form $(\exists x)\psi(x, a_1,\ldots,a_n)$, we argue as follows:
If $\mathcal{B}\models\exists\psi(x, a_1,\ldots,a_n)$, then there is an $a_{0}\in\mathrm{B}$ such that $\mathcal{B}\models\psi(a_{0}, a_1,\ldots,a_n)$ which is a quantifier-free formula; therefore, $\mathcal{A}\models\exists\psi(x, a_1,\ldots,a_n)$. Hence, $\mathcal{A}\models\psi(a_{0}, a_1,\ldots,a_n)$. We have shown that
$$\mathcal{B}\models\exists\psi(x, a_1,\ldots,a_n)\implies\mathcal{A}\models\exists\psi(x, a_1,\ldots,a_n)$$
Now, we intend to show that
$$\mathcal{A}\models\exists\psi(x, a_1,\ldots,a_n)\implies\mathcal{B}\models\exists\psi(x, a_1,\ldots,a_n)$$
If $\mathcal{A}\models\exists\psi(x, a_1,\ldots,a_n)$, then there is an $a_{0}\in\mathrm{A}$ such that $\mathcal{A}\models\psi(a_{0}, a_1,\ldots,a_n)$ which is a quantifier-free formula. Therefore, $\mathcal{B}\models\psi(a_{0}, a_1,\ldots,a_n)$. Hence, $a_{0}\in\mathcal{B}$, whereby we obtain the consequent.
Best Answer
This is really an instance of a more general fact.
To start, notice that for uncountable regular $\kappa$, $H_\kappa$ gets countability facts correctly. That is, if $p$ is countable, then $H_\kappa\vDash \exists f:\omega\to p \text{ surjective}$. Since $\omega, p\in M$, this sentence is true in $M$ as well. We now show that for any $q\in p$, we have also $q\in M$.
First fix $f_p\in M$ such that $M\vDash f_p:\omega\to p \text{ is surjective}$. Since $M$ thinks $f_p$ is a function, by elementary $H_\kappa$ thinks $f_p$ is a function, so $f_p$ is really a function. Now let $q\in p$ be arbitrary. We want to show that $q$ is an element of $M$. In $V$, we know that there is some $n$ such that $f_p(n)=q$. Also, for each $n\in\omega$, $n\in M$ by definability of the natural numbers. So we can ask for what $M$ thinks is $f_p(n)$. I claim that this object is $q$.
To see this, notice that $H_\kappa\vDash (\exists! x)(x\in p\wedge x=f_p(n))$. By elementarity, we have $M \vDash (\exists! x)(x\in p\wedge x=f_p(n))$. Now if $M$ is transitive then we are done: the unique $x$ is just $q$. But $M$ might not be transitive. We work around this obstacle by appealing to elementarity.
Fix the element $z\in M$ such that $M \vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. Passing this up to $H_\kappa$ by elementarity, we have $H_\kappa\vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. But then $H_\kappa$ also thinks that $q$ is the unique element satisfying the description. Hence $H_\kappa\vDash q=z$, and so $q$ and $z$ are really identical. This proves $q\in M$.
As to the question: notice that we have in general $(L_\kappa=H_\kappa)^L$. This means the general fact applies in this case, if we apply it in $L$. So just take $N_{\alpha+1}$ to be $M$ and take $p$ to be $N_\alpha$ and apply the fact.