Why is $\exists y \in B, \ \forall x \in A, \ P(x,y) $ stronger than $\forall x \in A,\ \exists y \in B, \ P(x,y)$

If general, a statement $F$ is told stronger than another statement $G$ if the implication $F \Rightarrow G$ holds (i.e. if $G$ holds whenever $F$ holds) but the converse implication $G \Rightarrow F$ does not hold (i.e. it is possible that $G$ holds but $F$ does not hold).

As you said, this is the case for $F = \exists y \in B \, \forall x \in A \, P(x,y)$ and $G = \forall x \in A \, \exists y \in B \, P(x,y)$, because if you assume $F$ then you can always prove $G$ (independently from the meaning of $A$, $B$ or $P$) but clearly the converse is not true: indeed, in the situation where $A = B = \mathbb{N}$ and $P = \, <$, we have that $F = \exists y \in \mathbb{N} \, \forall x \in \mathbb{N} : x < y$ is false ($\mathbb{N}$ has no maximum) but $G = \forall x \in \mathbb{N}\, \exists y \in \mathbb{N} : x < y$ (for every natural number $x$, its successor $x+1$ is greater than $x$).

It is stronger in the sense that anything you can prove with $\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$, can also be proven with $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$ because the latter implies the former. As an analogy: Every nut you can crack with a rubber hammer can also be cracked with a sledge hammer. But there might be nuts that can only be cracked with a sledge hammer. Therefore the sledge hammer is stronger than the rubber hammer.

Suppose $$ \exists y\in B,\forall x\in A,P(x,y). $$

Let $x_0\in A$. According to the previous assertion, there exists $y\in B$ such that for all $x\in A$, $P(x,y)$. In particular for $x=x_0$, we have $P(x_0,y)$. We just proved that for all $x_0\in A$, there exists $y\in B$ such that $P(x_0,y)$, which can be rewritten $$ \forall x_0\in a,\exists y\in B,P(x_0,y). $$

Of course you can replace $x_0$ with $x$, which gives $$ \forall x\in A,\exists y\in B,P(x,y). $$

Therefore, $$ \left( \exists y\in B,\exists x\in A,P(x,y)\right)\implies\left(\forall x\in A,\exists y\in B,P(x,y)\right). $$

The left-hand side is stronger in the sense that it implies the right-hand side. If you want to prove the right-hand side, it is succifient to prove the left-hand side, but it is not necessary. If you prove the right-hand side by proving the left-hand side, we consider that you prove a stronger result.