Now, I have been reading the book. I summarize
Axiom 1 (of Separation). Let $A$ a set, and for each $x \in A$, let $\varphi(x)$ a property pertaining to $x$. Then there exists a set $C := \{x \in A : \varphi(x) \text{ is true} \}$ (or $\{x \in A : \varphi(x)\}$ for short), whose elements are precisely the elements $x$ in $A$ for which $\varphi(x)$ is true.
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ).$$
Axiom 2 (of Extensionality). Two sets $A$ and $B$ are equal, $A = B$, if every element $x$ of $A$ belongs also to $B$, and every element $y$ of $B$ belongs also to $A$.
$$\forall x \; (x \in A \iff x \in B) \implies A = B.$$
Your theorem
Theorem 3. Let $A$ and $B$ be sets. Then exists a unique set $C$ whose elements belongs to both $A$ and $B$.
$$\exists ! C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B).$$
Now, to prove the theorem we need show the existence and uniqueness. First the existence. Let $A$ and $B$ be sets, and let $\varphi(x) := x \in B$. Then, by Axiom 1, we have
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ),$$
i.e.,
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B ).$$
This prove the existence of the intersection set $C$ for any sets $A, B$.
We now show the uniqueness. Let $A$ and $B$ sets. Suppose there exists two sets $C, C'$ such that
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B )$$
and
$$\exists C' \, \forall x \; (x \in C' \iff x \in A \;\land\; x \in B ).$$
Using the notation, we have
$$x \in A \;\land\; x \in B \iff x \in C',$$
i.e.,
$$x\in C \iff x \in A \;\land\; x \in B \iff x \in C'.$$
This means, the statements $x\in C$, $x \in A \;\land\; x \in B$, and $x \in C'$ are equivalents. Thus, we have
$$x \in C \iff x \in C'.$$
By Axiom 2, we have $C = C'$ as desired. $\;\Box$
Now, note that to use the Axiom 1, you need two sets: a reference set from which construct the intersection, and another to define the property. Also, another valid definition of intersection is
$$A \cap B = \{x \in A : x \in B\}.$$
So the set $z$ that you mention is $A$, and your property $x \in A \; \land \; x \in B$ should be just $x \in B$.
Finally, in Set theory, the Axiom of union exists because this does not follow from the other axioms (empty set, extensionaity, separation, etc). These axioms allow us to build smaller sets from other reference sets. But the union set is a larger set from anothers, because of that this axiom is necessary.
I have now figured this out, We need the power set axiom introduced in the next section as indicated by the Hint in the Dover edition (which I am not using) this hint is not in the original edition.
From this we can deduce. $\exists Z \forall x(x \in Z \iff \exists C(x= A \cap C \space \& \space C \in B)$ as per the definition by abstraction, and Whence the desired inference. I knew something wrong was going on here in the edition I have.
Best Answer
The author is defining the "set-builder" operator: $\{ x \mid \varphi(x) \}$ that "maps" a predicate (a formula $\varphi$ with free variable $x$) into a term (i.e. the "name" of an object).
It is well-known that the so-called unrestricted Comprehension Principle is inconsistent [see §1.3]: thus, not every predicate can meaningfully define a set.
The author uses the definitional schema illustrated at page 19 with the $x/y=z$ example.
Going back to Definition 11, the author illustrates it at page 34.
There are two possible cases:
in which case we define that the "set-builder" operator maps formula $\varphi$ to that set, or
in which case we "arbitrarily" define that the "set-builder" operator maps formula $\varphi$ to the empty set.
Maybe your confusion is due to the incorrect way of reading the definition:
We are defining $y$, i.e. we have to start from the formula and "manufacture" the corresponding set.
Regarding:
The issue is not the satisfiability of the formula; consider the discussion about Russell's paradox (page 6).
Into the formula $(\forall x) (x \in y \leftrightarrow \lnot (x \in x))$we are using $\lnot (x \in x)$ as $\varphi(x)$ and the formula is indeed satisfiable: $\lnot (\emptyset \in \emptyset)$.