Maybe something is missing in the original statement and is: whose kernel is that on the right hand side?
First of all, let's recall that $f$ and $g$ are maps from $X$ to $Y$.
Then that kernel on the right is obviously
$$
\ker (f_{*}, g_{*}) \subset \mathrm{Hom}(Z,X) \ ,
$$
where
$$
f_{*}, g_{*} : \mathrm{Hom}(Z,X) \longrightarrow \mathrm{Hom}(Z,Y)
$$
are the maps defined by $\varphi \mapsto f\circ \varphi$ and $\varphi \mapsto g\circ \varphi$, respectively.
That said, the bijection you ask for simply sends every map
$$
\varphi : Z \longrightarrow \ker (f,g) \subset X
$$
to $i\circ \varphi$, where $i: \ker (f,g) \hookrightarrow X$ is the inclusion.
As for the other direction:
$$
\psi \in \ker (f_{*}, g_{*}) \ \Longleftrightarrow \ f\circ \psi = g\circ \psi \ \Longleftrightarrow \ f(\psi (z)) = g(\psi (z)) \ \text{for all} \ z \in Z
$$
Which means:
$$
\psi (z) \in \ker (f,g) \ \text{for all} \ z \in Z \ \Longleftrightarrow \ \mathrm{im} (\psi ) \subset \ker (f, g) \ .
$$
So, $\psi : Z \longrightarrow X$ factorises through the inclusion $i: \ker (f,g) \hookrightarrow X$. That is, since $i$ is injective, there exists a unique $\varphi : Z \longrightarrow \ker (f,g)$ such that $\psi = i \circ \varphi$.
Hence, the map $LHS \longrightarrow RHS$ sends $\varphi $ to $i \circ \varphi$ and the map $RHS \longrightarrow LHS$ sends $\psi$ to $\varphi$ such that $\psi = i \circ \varphi$.
Both compositions are the identity by definition:
$$
\psi \mapsto \varphi \mapsto i \circ \varphi = \psi
$$
and
$$
\varphi \mapsto i \circ \varphi \mapsto \varphi \ .
$$
Conclusion: I've never seen "two" sets that were so much the same. :-)