It's perhaps best to think about this first in the context of a particular structure and a particular formula, where you can easily see what's going on. Consider the structure of the natural numbers with the usual ordering relation $<$. The sentence $\forall x\exists y\,(x<y)$ is true in this structure, because for every natural number $x$ there is a larger one $y$, for example $x+1$ (or $x+17$ or $(2x)!$ or lots of other choices). Equivalently, the second-order sentence $\exists F\forall x\,(x<F(x))$ is true in the same structure, witnessed by functions like $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$).
On the other hand, we cannot say (as you did) that $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. To see this, notice first that this "logical consequence" relationship concerns structures for a language that has not only the relation symbol $<$ but also the function symbol $F$. Consider the structure of natural numbers with $<$ as before and some interpretation of the function symbol $F$. Then $\forall x\exists y\,(x<y)$ is true (regardless of the interpretation of $F$ because it doesn't mention $F$), but whether $\forall x\,(x<F(x))$ is true depends on how we interpret $F$. It's true if we take $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$), but it's false if we take $F(x)=0$ (or $F(x)=17$ or $F(x)=\lfloor x/2\rfloor$, or lots of other functions). But $\models$ means that every structure satisfying what's on the left of $\models$ also satisfies what's on the right. So in our case, it would require that $\forall x\,(x<F(x))$ is true regardless of how we interpret $F$ as long as $\forall x\exists y\,(x<y)$ is true. Since that doesn't hold when $F$ is "badly" interpreted ($0$ or $17$ or $\lfloor x/2\rfloor$), we don't have $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. The most we can say is that, having satisfied $\forall x\exists y\,(x<y)$, we can, by suitably interpreting $F$, also satisfy $\forall x\,(x<F(x))$ --- and vice versa. That's what "equisatisfiable" means.
Yes, the first move you made does preserve equivalence, and here are some basic rules for pulling a quantifier outside logical operators, where P does not contain any free variables x:
$P \lor \forall x \: \phi(x) \equiv \forall x (P \lor \phi(x))$
$P \lor \exists x \: \phi(x) \equiv \exists x (P \lor \phi(x))$
$P \land \forall x \: \phi(x) \equiv \forall x (P \land \phi(x))$
$P \land \exists x \: \phi(x) \equiv \exists x (P \land \phi(x))$
$P \rightarrow \forall x \: \phi(x) \equiv \forall x (P \rightarrow \phi(x))$
$P \rightarrow \exists x \: \phi(x) \equiv \exists x (P \rightarrow \phi(x))$
$\forall x \: \phi(x) \rightarrow P \equiv \exists x (\phi(x) \rightarrow P)$
$\exists x \: \phi(x) \rightarrow P \equiv \forall x (\phi(x) \rightarrow P)$
So take care of those last two: the quantifier changes if you pull it outside a conditional, and it is the antecedent of that conditional!
And because of the latter, there is no simple equivalence involving puling a quantifier outside a biconditional.
Also, you can move a universal over another universal (and same for existentials), but you can't move a universal over an existential or vice versa. That is:
$\forall x \forall y P(x,y) \equiv \forall y \forall x P(x,y)$
and
$\exists x \exists y P(x,y) \equiv \exists y \exists x P(x,y)$
But not:
$\forall x \exists y P(x,y) \equiv \exists y \forall x P(x,y)$
Best Answer
The answer to your general question is very much "no", because moving a quantifier past a connective can change the quantifier. For instance, $\neg \forall x \phi(x)$ is equivalent to $\exists x \neg\phi(x)$ and in the vast majority of cases is not equivalent to $\forall x \neg \phi(x)$.