$z \cap x$ denotes a set whose existence is proved using the axiom of replacement? I am reading the book Mathematical Logic and Model Theory by Prestel and Delzell and after they talked about first order semantics there was a section on the axiomatization of set theory. They present the foundation axiom in the following way:
$$\forall x (x\neq\emptyset \longrightarrow \exists z (z\in x \wedge z \cap x=\emptyset))$$
Then they go on to say that $z\cap x$ (for intersection) denotes a set whose existence is proved using the axiom of replacement and whose uniqueness is guaranteed by the axiom of extensionality. 
I don't get this at all. The more I think about it the more confused I am.They don't define $z\cap x$. Is the axiom supposed to be the definition of it or are we supposed to take the usual definition? 
Could someone explain how to prove existence and uniqueness of $z\cap x$ ?

In fact you (and they) don't need to use $\cap$ in order to state Regularity. As they haven't defined the symbol, they shouldn't use it. You can eliminate intersection simply by expanding its definition:
$$
\text{AxRegularity}\iff \forall x (x\neq\emptyset \longrightarrow \exists z (z\in x \wedge \forall u \neg (u\in z \wedge u\in x))
$$
Their statement about why $z\cap x$ exists is, frankly, a little murky. Given two sets $X,Y$, Separation guarantees existence of a set $Z$ such that
$$
u\in Z \longleftrightarrow (u\in X \wedge u\in Y) 
$$
i.e. $Z = \{u\in X\mid u\in Y\}$. Uniqueness of $Z$ follows by Extensionality. Of course, $Z$ is just $X\cap Y$.
You need Replacement to show that $\{z\cap x\mid z\in x\}$ exists, but it's not necessary to show that: all you need is existence of the intersection of any two sets.
