When you negate a quantifier, you 'bring the negation inside', e.g. $\neg \forall x P(x)$ is equivalent to $\exists x \: \neg P(x)$, where P(x) is some claim about $x$.
If you have two quantifiers, that still works the same way, e.g. $\neg \forall x \exists y P(x,y)$ is equivalent to $\exists x \neg \exists y P(x,y)$, which in turn is equivalent to $\exists x \forall y \neg P(x,y)$. And once you see that, you can understand that you can move a negation through a series of any number of quantifiers, as long as you change the quantifier: each $\forall$ becomes a $\exists$ and vice versa.
Also, since these are all equivalences, you can also bring negations outside, if that's what you ever wanted to, again as long as you change each quantifier that you move the negation through. For this reason, this is sometimes called the 'dagger rule': you can 'stab' a dagger (the negation) all the way through a quantifier, thereby changing the quantifier.
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
Disclaimer: I just had a quick look over the book you've mentioned. It seems that it is not dealing with logic in a pedantic mathematical way, so I'm a bit scared of confusing you even more. If so, you should ignore this answer.
Meta- vs. Object-language
As it happens often to me, I assume that the very core of your question comes from a confusion of meta-language with object-language. Think of (now I take my quick review on Suppes as a foundation -- it might have appeared in the book) of object-language as a certain instance of symbols plus all the logical symbols like $\rightarrow, \wedge, \forall$ etc. For example:
If we want to formalize this sentence, we need a constant $a = \texttt{this apple}$ and a relation $Gx = \texttt{x tastes good}$. Hence, our object-language is $L = (a, G)$ (I omit the logical symbols since they are part of any language).
So, a statement within the object-language is a formula using only logical symbols and $a$ or $G$, like $(\forall x) (Gx)$, $(Ga)\wedge(\exists x)(Gx)$, etc. And a proof within the object-language is a finite sequence of statements within the object-language where each statement of the sequence is either a so called assumption or is the conclusion from applying a rule of inference to (some of) the assumptions. Whatever: You infer only with statements written in the object-language.
On the other side, any statement which states something about languages in general (and therefore not in a certain language) is a statement in meta-language. Consider the statement you've already shown:
or clearer it states:
Now, this is clearly not a statement in any object-language. It is a statement about object-languages! For example about $L=(P,a,b)$ or $L=(Q)$ as long as $P,Q$ are binary relations(symbols).
So, actually, the statement is:
[Side-kick one could ignore: Of course, one can formalize the meta-language, so it becomes object-language but "one level up". There has to be another meta-language to formalize the former meta-language -- and anyway it is still the meta-language for the former object-language. So: these aren't absolute but relative terms. But for now, consider them as absolute in the above sense.]
Back to the question
Okay, so far, now back to your question: What I would say what has brought up your question is a confusion about meta- and object-language.
In fact, the statement you want to prove within your question is a statement in meta-language as it states something about languages in general. Here: About logical truths.
The point is that within the meta-language you know about natural numbers and all that stuff. Actually, you know about set-theory. It's like "usual mathematics", it's the language mathematicians use to proof things plus a lot of everyday language (at least in our context here, if you've read the side-kick).
Your statement in question is something like:
So, feel free to use the natural numbers (and therefore induction and so on) to prove this.