So you have $φ$ and want to show continuity of
$$F:Z×X\to Y\\ (z,x)\mapsto φ(z)(x)$$
The idea is to write $F$ as the composition of two maps, namely
$$Z\times X\to C(X,Y)×X\to Y$$
where the first map is $\varphi×1_X$ which is continuous by assumption, and the second map is the evaluation $$ε:C(X,Y)×X\to Y\\ (f,x)\mapsto f(x)$$
Now $ε$ is continuous, and the proof goes as follows:
Take a point $(f,x)$ where $f:X\to Y$ and $x\in X$, and consider an open neighbourhood $V$ of $ε(f,x)=f(x)$. By continuity of $f$, there is an open neighbourhood $U$ of $x$ such that $f[U]⊆V$. By local compactness of $X$, there is also a compact neighbourhood $K$ with $x\in K⊆U$. Then $(K,V)×K$ is a neighbourhood of $(f,x)$ such that $ε(g,y)=g(y)\in V$ for $g\in(K,V)$ and $y\in K$. Hence $ε$ is continuous.
There is a converse to this: Given a continuous $F:Z×X\to Y$, then
$$\hat F:Z\to C(X,Y)\\ \hat F(z)(x)=F(z,x)$$
is continuous. To show this, let $z\in Z$ and $(K,V)$ a subbase neighborhood of $\hat F(z)$. Then $\hat F(z)(x)\in V$ for all $x\in K$, thus
$\{z\}×K⊆F^{-1}[V]$ which is open. By the Tube Lemma we can find an open $U$ such that $\{z\}×K⊆U×K⊆F^{-1}[V]$. Then $U$ is a neighbourhood of $z$ such that $\hat F[U]⊆(K,V)$.
Furthermore, now that we have this $\hat F$ we can consider
$$Z×X\xrightarrow{\hat F×1_X}C(X,Y)×X\xrightarrowεY$$ and we see that this gives us our original $F$, which means that turning $F$ into $\hat F$ is sort of a converse to turning a $φ:Z\to C(X,Y)$ into $F$. After all we get a bijection
$$C(Z,C(X,Y))\cong C(Z×X,Y)$$
To get a picture of how a continuous map $\phi:Z\to C(X,Y)$ looks like, you could take as an example the case $X=Y=Z=\Bbb R$. A map $\Bbb R\to\Bbb R$ can be thought of as an unbroken curve without "jumps". A function $\Bbb R\to C(\Bbb R,\Bbb R)$ assigns to each real number $z$ such a curve, let's denote the function assigned to $z$ by $f_z$. We can put these $f_z$ side by side, so that the graph of $f_z$ "hovers" over the line $\{z\}×\Bbb R$. All those maps together give a function $\Bbb R×\Bbb R\to\Bbb R$, namely the function $(z,x)\mapsto f_z(x)$. There is no reason why $F$ should be continuous if $φ$ is merely a function, but the topology on $C(\Bbb R,\Bbb R)$ allows us to demand a continuous $φ$. Intuitively, $φ$ is continuous if for small changes in $z$, the maps $f_z$ do not change drastically. It turns out that the compact-open topology is the suitable one. I invite you to figure out what a continuous $φ$ means if the topology on $C(\Bbb R,\Bbb R)$ is the product topology. Actually, in that case this makes the map $F$ only continuous in both coordinates $z$ and $x$, but not continuous overall.
