Let's try to express $(\forall x, \exists y, (P(x) \rightarrow Q(y))) \to (\exists y, \forall x, (P(x) \rightarrow Q(y)))$ in English:
Premise: suppose that for every $x$, there is a $y$ such that $P(x)$ implies $Q(y)$.
Conclusion: then there is a $y$ such that for every $x$, $P(x)$ implies $Q(y)$.
Suppose the premise holds, and suppose that we have found an $x_0$ such that $P(x_0)$ holds. Then there is a $y_0$ such that $Q(y_0)$ holds. There is nothing specific to $x_0$ about $y_0$: if there is some other $x$ that satisfies $P(x)$, then we can take that same $y_0$, we still have $Q(y_0)$, hence we still have that $P(x)$ implies $Q(y)$.
Or, in other words, if the premise holds and there is an $x_0$ such that $P(x_0)$ holds, then the conclusion holds. We have proved that the formula is always true under the additional assumption $\exists x_0, P(x_0)$.
So all we need to do now is look at what happens if that additional assumption does not hold. By the principle of excluded middle and the well-known principles of inversion of negation and quantifiers, the opposite assumption is $\forall x, \neg P(x)$. Since $P(x)$ never holds, the implication $P(x) \rightarrow Q(y)$ always holds. Are we finished? Almost. If we take any $y_0$ in the model, then the formula $\forall x, (P(x) \rightarrow Q(y_0))$ holds. So the conclusion is true in every non-empty model, hence the formula holds in every non-empty model.
If you allow an empty model, then the formula does not hold, because the premise is true ($\forall x, …$ is vacuously true) and the conclusion is false ($\exists y, …$ is vacuously false).