Apart from performing a somewhat lengthy diagram chase, the only thing to do here is to notice that giving a map $(a,b) : A \to [K,X] \mathrel{\times_{[K,Y]}} [L,Y]$ amounts to giving a commutative diagram as on the left by the universal property of the pull-back $\DeclareMathOperator{\Hom}{Hom}$

while the diagram on the right is equivalent to it by the adjunction between the cartesian product and the internal $\Hom$ and the definitions of $p_\ast$ and $i^\ast$.
Let me give names to the maps in the first commutative diagram of the proof you ask about:

The map $f: \Lambda_{k}^n \to [L,X]$ corresponds to a map $\tilde{f}: \Lambda_{k}^n \times L \to X$ while giving the bottom map $(g,h) : \Delta^n \to [K,X] \mathrel{\times_{[K,Y]}} [L,Y]$ amounts to either of the two commutative diagrams

Using the universal property of the pull-back defining $[K,X] \mathrel{\times_{[K,Y]}} [L,Y]$ we see that asserting the commutativity of the square we started with is equivalent to giving the map $f: \Lambda_{k}^n \to [L,X]$, the commutative square $(1)$ and requiring the two squares

to be commutative.
Passing to the adjoint side using the map $\tilde{f} : \Lambda_{k}^n \times L \to X$ and the square $\widetilde{(1)}$, the commutativity of the squares $(2)$ and $(3)$ is equivalent to the two commutative squares

At this point I think I can leave it to you to contemplate the commutative diagrams $\widetilde{(1)}$, $\widetilde{(2)}$ and $\widetilde{(3)}$ and the push-out square defining $(\Lambda_{k}^n \times L) \mathrel{\cup_{(\Lambda_{k}^n \times K)}} (\Delta^n \times K)$ and to think about what it means to define a map from $(\Lambda_{k}^n \times L) \mathrel{\cup_{(\Lambda_{k}^n \times K)}} (\Delta^n \times K)$ in order to see that giving the squares $\widetilde{(1)}$, $\widetilde{(2)}$ and $\widetilde{(3)}$ and requiring their commutativity is equivalent to giving the commutative square

as claimed by Goerss and Jardine.