Where in the proof of this theorem shows "If (x, y)$\in f$ and (x, z) $\in f$, then y=z."? 
Definition 8. Let X and Y be sets. A function from X to Y is a triple (f, X, Y), where f is a relation from X to Y satisfying
  (a) Dom(f) = X.
  (b) If (x, y)$\in f$ and (x, z) $\in f$, then y=z.
  We shall adhere to the custom of writing f: $X\space \rightarrow Y$ instead of (f, X, Y) and $y=f(x)$ instead of $(x,\space y) \in f$.
  Source: Set Theory You-Feng Lin, Shwu-Yeng T.Lin  
Theorem 8 Let $f: A \rightarrow C$ and $g: B\rightarrow D$ be two functions such that $f(x)=g(x)$, $\forall x\in A \cap B$. Then the union h of f and g defines the function
       h =$f\cup g$: $A \cup B \rightarrow C \cup D$
  where    h = $\begin{cases}
f(x),  & \text{if } x \in A \\
g(x), & \text{if } x \in B
\end{cases}$
  [Proof] Since $f$ and $g$ are relations, $f⊆A×C$ and $g⊆B×D$, and we have
   $$h​=f∪g ⊆ (A×C)∪ (B×D)$$
   $$ ⊆ (A∪B)× (C∪D)  $$
  because both A×C and B×D are subsets of (A∪B)×(C∪D).
  Thus h is a relation from A∪B to C∪D. We leave it to the reader to verify that
   $$Dom(h)=Dom(f)∪Dom(g)$$
          $$=A∪B$$.
  This shows that the relation $h$ satisfies Definition 8(a)
    For each element $x \in A \cup B$, we may consider the following three cases:
    $$(1) x \in A-B, (2) x \in B-A, \text{and (3)} x \in A \cap B $$
    Since $f: A \rightarrow C$ and $g: B \rightarrow D$ satisfy Definition 8(b) and $f(x) = g(x)$ $\forall x$ $\in A \cap B$, we have that h(x) is uniquely defined in each of the three cases. Thus, the relation h satisfies Definition 8(b) as well. Hence, 
  $$h: A\cup B \rightarrow C\cup D$$
  is indeed a function.

I see nowhere in the proof showing justification for $f: A \rightarrow C$ and $g: B \rightarrow D$ satisfying Definition 8(b). "(b) If (x, y)$\in f$ and (x, z) $\in f$, then y=z." Can you explain how it leads to the satisfaction of the Definition 8(b)?

Just because $f$ is a function, hence $y=f(x)$ and $z=f(x)$ must be equal,
otherwise $f$ cannot be a function.
