Quantifier order is not commutative

Forward direction. Assume $\mathcal{M} \models_s \exists y\, \forall x\, R(x,y)$. By the semantic clause for $\exists$, $\mathcal{M} \models_s \exists y\, \varphi \iff \text{there is } d \in M \text{ with } \mathcal{M} \models_{s[y \mapsto d]} \varphi$; applying it here gives some $d_0 \in M$ with $\mathcal{M} \models_{s[y \mapsto d_0]} \forall x\, R(x,y)$.

By the clause for $\forall$ applied to this, $\mathcal{M} \models_{s[y \mapsto d_0][x \mapsto a]} R(x,y)$ holds for every $a \in M$, since the $\forall x$ ranges over all of $M$ regardless of which $a$ we later pick.

Now fix an arbitrary $a \in M$ to verify $\forall x\, \exists y\, R(x,y)$. From the previous paragraph, taking $x \mapsto a$ gives $\mathcal{M} \models_{s[y \mapsto d_0][x \mapsto a]} R(x,y)$, which exhibits $d_0$ itself as a witness for $\exists y\, R(x,y)$ under $x \mapsto a$; hence $\mathcal{M} \models_{s[x \mapsto a]} \exists y\, R(x,y)$.

Since $a \in M$ was arbitrary, $\mathcal{M} \models_{s[x \mapsto a]} \exists y\, R(x,y)$ holds for every $a$, which is exactly the semantic clause for $\mathcal{M} \models_s \forall x\, \exists y\, R(x,y)$. This proves $\exists y\, \forall x\, R(x,y) \Rightarrow \forall x\, \exists y\, R(x,y)$.

Converse fails: a counterexample. Let $\mathcal{M}$ have domain $M = \mathbb{Z}$ (the integers) and interpret $R(x,y)$ as "$x < y$". Then $\forall x\, \exists y\, R(x,y)$ is TRUE: for every integer $x$, taking $y = x+1$ gives $x < y$. But $\exists y\, \forall x\, R(x,y)$ is FALSE: it would require a single integer $y$ larger than every integer $x$, and no such maximum integer exists (for any candidate $y$, the integer $y+1$ violates $y+1 < y$). So $\forall x\, \exists y\, R(x,y)$ holds while $\exists y\, \forall x\, R(x,y)$ fails, showing the converse implication is false in general.