Let $S$ be a set of (logical) formulae and $\psi$ be a formula. Then $S \vdash \psi$ means that $\psi$ can be derived from the formulae in $S$. Intuitively, $S$ is a list of assumptions, and $S \vdash \psi$ if we can prove $\psi$ from the assumptions in $S$.
$\vdash \psi$ is shorthand for $\varnothing \vdash \psi$. That is, $\psi$ can be derived with no assumptions, so that in some sense, $\psi$ is 'true').
More precisely, systems of logic consist of certain axioms and rules of inference (one such rule being "from $\phi$ and $\phi \to \psi$ we can infer $\psi$"). What it means for $\psi$ to be 'derivable' from a set $S$ of formulae is that in a finite number of steps you can work with (i) the formulae in $S$, (ii) the axioms of your logical system, and (iii) the rules of inference, and end up with $\psi$.
In particular, if $\vdash \psi$ then $\psi$ can be derived solely from the axioms by using the rules of inference in your logical system.
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:
This apple tastes good.
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:
For every binary relation(symbol) $P$ it holds that
\begin{align*}\models (\forall x)(\forall y)(Pxy) \rightarrow (\forall y)(\forall x)(Pxy)\end{align*}
or clearer it states:
For every binary relation(symbol) $P$ it holds that one can infer $(\forall x)(\forall y)(Pxy) \rightarrow (\forall y)(\forall x)(Pxy)$ without any assumptions.
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:
For every language $L$ and every relation (symbol) $R$ within $L$, it holds that ...
[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:
For every $n\in\mathbb{N}\setminus \{0\}$, every bijective $\sigma : \{1,\ldots,n\} \rightarrow \{1,\ldots,n\}$, every language $L$ and every $n$-ary relation(symbol) R within the language $L$ it holds that
\begin{align*}\models (\forall x_1) \ldots (\forall x_n) (Rx_1\ldots x_n)\rightarrow (\forall x_{\sigma(1)})\ldots (\forall x_{\sigma(n)}) (Rx_1\ldots x_n)\end{align*}
So, feel free to use the natural numbers (and therefore induction and so on) to prove this.
Best Answer
Yes, it is a quantifier. You can write the equivalence
$$\not\exists a:p(a)\equiv\forall a:\lnot p(a).$$