Using the regularity of $\kappa$ and the fact that the infinite cardinal $\lambda$ is less than $\kappa$, Jech notes that
$$ \kappa^\lambda=\bigcup_{\alpha<\kappa}\alpha^\lambda, $$
where both sides are understood as sets of functions. From this, he claims that, as cardinals,
$$ \kappa^\lambda=\sum_{\alpha<\kappa}|\alpha|^\lambda. $$
This would be obvious if the sets $\alpha^\lambda$ were disjoint as $\alpha$ varies but, as you point out, this is clearly not the case.
You suggest instead to replace these sets by disjoint copies, say replacing $\alpha$ by $\hat\alpha=\{\alpha\}\times\alpha$ for each $\alpha<\kappa$. This changes the union $\bigcup_\alpha\alpha^\lambda$ into the disjoint union $\bigcup_\alpha\hat\alpha^\lambda$. Note that there is an obvious injection from the former into the latter: Given any function $f\in\bigcup_\alpha\alpha^\lambda$, find the least $\alpha$ such that $f\!:\lambda\to\alpha$, and map $f$ to its copy in $\hat\alpha^\lambda$.
This means that
$$ \kappa^\lambda\le\sum_{\alpha<\kappa}|\alpha|^\lambda. $$
However, Jech is claiming more, namely, equality rather than the inequality we just showed.
Luckily for us, the inequality
$$ \sum_{\alpha<\kappa}|\alpha|^\lambda\le\kappa^\lambda $$
is quite easy to establish: First, for each $\alpha<\kappa$, $|\alpha|^\lambda\le\kappa^\lambda$, so $\sum_{\alpha<\kappa}|\alpha|^\lambda \le \sum_{\alpha<\kappa}\kappa^\lambda=\kappa\cdot\kappa^\lambda=\kappa^\lambda$. Note that this is the only place where we used that $\lambda$ is infinite (in fact, it suffices that $\lambda>0$).
From the two inequalities, the claimed equality now follows. Let me close by pointing out that this is a really useful heuristic: just as equality of sets is two containments, many equalities in cardinal arithmetic are really two inequalities.
In the case at hand, one of the inequalities turned out to be obvious upon inspection. This is actually not so uncommon. The strategy you suggested, of replacing a union by a disjoint union so that its cardinality can be estimated via Jech's definition of infinite sums of cardinals, turns out to be very useful in practice.
Alright, I'll answer the question. Yes, I found out that we don't need $f$ in this proof. Also, the one thing I didn't explain in the question is why $C$ is a Club set. It follows from the fact that $C$ is the diagonal intersection of the sets $C_{\alpha}=\{\delta<\aleph_1: a_{\alpha}\subseteq\delta\}$ which are all obviously Club sets. This is not the shortest proof the $\Delta$-System lemma, but I think it's still a nice proof.
Best Answer
This is because the intersection of a stationary set and a club is stationary.
Suppose $S$ is stationary and $C$ is a club. Let $C′$ be any club. Then $C\cap C′$ is a club. Since $S$ is stationary, $S\cap(C\cap C′)$ is nonempty. Hence $(S\cap C)\cap C′$ is nonempty. Since $C′$ is an arbitrary club, it follows that $S\cap C$ is stationary.