The sets $A_\alpha$ are defined in terms of the function $f$, so you certainly can’t ‘work directly with the elements of $A_\alpha$’ without defining $f$ first. Since that looks like a fairly fundamental misunderstanding, I’m not sure just where your problem really lies, so I’m just going to add a lot of extra explanation to Jech’s argument in hopes that some of it will help.
Let’s be clear first on exactly what’s we’re trying to do here. I’m going to change Jech’s notation a little to try to make matters clearer.
We’re starting with a sequence $\langle \mathcal{S}_\alpha:\alpha<\omega_1\rangle$ such that each $\mathcal{S}_\alpha$ is a countable collection of subsets of $\alpha$. This sequence has the property that if $X$ is any subset of $\omega_1$, $\{\alpha<\omega_1:X\cap\alpha\in \mathcal{S}_\alpha\}$ is stationary. From this we want to construct a $\diamondsuit$-sequence $\langle S_\alpha:\alpha<\omega_1\rangle$: each $S_\alpha$ is a single subset of $\alpha$, and if $X$ is any subset of $\omega_1$, then $\{\alpha<\omega_1:X\cap\alpha=S_\alpha\}$ is stationary. In other words, we want to reduce the countable families $\mathcal{S}_\alpha$ to single sets $S_\alpha$. The coding using $f$ is a mechanism for doing this.
The basic idea behind $f$ is make $\omega_1$ ‘look like’ $\omega$ copies of itself. More, we want to do this in such a way that $f$ makes each limit ordinal $\alpha$ ‘look like’ $\omega$ copies of itself. That’s why we take $f$ to be a bijection $f:\omega_1\to\omega_1\times\omega$ with the additional property that $f“\alpha=\alpha\times\omega\,$ for each limit $\alpha<\omega_1$. We can’t guarantee this nice property for every $\alpha<\omega_1$, but we can get it for all limit ordinals. And it’s enough to work with limit ordinals, since the set of limit ordinals is a cub: anything that happens on a stationary subset of $\omega_1$ happens on a stationary set of limit ordinals, and vice versa.
Now for each limit ordinal $\alpha<\omega_1$ let $\mathcal{A}_\alpha=\{f“S:S\in\mathcal{S}_\alpha\}$. Each $S\in\mathcal{S}_\alpha$ is a subset of $\alpha$, so by the choice of $f$ we know that $f“S\subseteq f“\alpha=\alpha\times\omega$. Thus, the map $f$ has spread each $S\in\mathcal{S}_\alpha$ out over $\alpha\times\omega$, and $\mathcal{A}_\alpha$ is a countable collection of subsets of $\alpha\times\omega$. Because it is countable, we can index $\mathcal{A}_\alpha=\{A_\alpha^n:n\in\omega\}$, where each $A_\alpha^n\subseteq\alpha\times\omega$. These sets $A_\alpha^n$ are just the images under $f$ of the members of $\mathcal{S}_\alpha$.
Now I’m going to chop up the sets $A_\alpha^n$: for each limit $\alpha<\omega_1$ and each $n\in\omega$ let $S_\alpha^n=\{\xi<\alpha:\langle\xi,n\rangle\in A_\alpha^n\}\;.$ If you draw a sketch of $\alpha\times\omega$, you can visualize $S_\alpha^n$ as the projection on the first factor of $A_\alpha^n\cap(\alpha\times\{n\})$. For each $n\in\omega$ we’re intersecting $A_\alpha^n$ with $\alpha\times\{n\}$ and projecting to the first factor; in a sense we’re taking just a horizontal slice of $A_\alpha^n$, and we’re slicing at a different ‘level’ for each member of $\mathcal{A}_\alpha$. The claim is that one of the $\omega$ slices ‘works’, in the sense that there is at least one $n\in\omega$ such that $\langle S_\alpha^n:\alpha<\omega_1\rangle$ is a $\diamondsuit$-sequence.
If not, then for each $n\in\omega$ there must be a ‘bad’ subset $X_n$ of $\omega_1$, ‘bad’ in the sense that the set $B_n=\{\alpha<\omega_1:X_n\cap\alpha=S_\alpha^n\}$ is non-stationary. Let $X=\bigcup_{n\in\omega}\Big(X_n\times\{n\}\Big)\subset\omega_1\times\omega\;;$
it follows immediately from the definition of $\mathcal{A}_\alpha$ that $\{\alpha<\omega_1:X\cap(\alpha\times\omega)\in\mathcal{A}_\alpha\}$ is stationary. (This is true for any subset of $\omega_1\times\omega$, not just this $X$.) Since a countable union of non-stationary sets is non-stationary, there must be some $n\in\omega$ such that $T=\{\alpha<\omega_1:X\cap(\alpha\times\omega)=A_\alpha^n\}$ is stationary.
Suppose that $\alpha\in T$. Then $X\cap(\alpha\times\omega)=A_\alpha^n=S_\alpha^n\times\{n\}\subseteq\omega_1\times\{n\}\;,$ and $X\cap(\omega_1\times\{n\})=X_n\times\{n\}$, so $(X_n\cap\alpha)\times\{n\}=X\cap(\alpha\times\{n\})=S_\alpha^n\times\{n\}\;,$ and $X_n\cap\alpha=S_\alpha^n$. But then $T\subseteq B_n$, and $B_n$ is stationary, which is a contradiction.