What is the definition of a labeled function? I always see that people label their functions by giving an index. Specifically I have this example: 
$Theorem$: There is a unique binary operation $+:\mathbb{N}\times\mathbb{N}$ that satisfies the following two propierties for all $n,m\in N$
$1)$ $n+1=s(n)$
$2)$ $n+s(m)=s(n+m)$
Note: funcion $s$ is given
$proof (existence):)$ Let $p\in \mathbb{N}$. By the theorem of recursion, there is a unique function $f_{p}: \mathbb{N}\longrightarrow \mathbb{N}$ such that $f_{p}(1)=s(p)$ and $f_{p}\circ s=s\circ f_{p}$. Let $+:\mathbb{N} \longrightarrow \mathbb{N}$ be defined by$ c+d=f_{c}(d)$ for all $c,d\in \mathbb{N}$. Let $n,m\in \mathbb {N}$. Then $n+1=f_{n}(1)=s(n)$, which is part $(1)$, and $n+s(m)=f_{n}(s(m))=(f_{n}\circ s)(m)=(s\circ f_{n}(m))=s(f_n(m))=s(n+m)$, which is part $(2)$
So, here I'm confused because I don't know what the subindex $p$ means. I can see it like 1) a name for the function ( and we are talking about a universal instantiation) or 2) I can see it like a parameter (and then $f_{p}:\mathbb{N}\longrightarrow \mathbb{N}$ stands for a particular case of $f:\mathbb{N}\times\mathbb{N}\longrightarrow \mathbb{N}$). 
I hope you can see my point. Another examples: "Let $n\in \mathbb{N}$ and define for all $x\in \mathbb{R}$ $f_{n}(x)=(n+1)x^{2}$ " Does this means $f:\mathbb{N}\times  \mathbb{R}\longrightarrow \mathbb{R}$?.
Yet another example:  "Let $X$ be a set. We talk about the identity function of $X$ to be defined as $I_{X}(y)=y$ for every $y\in X$". 
In general, what is the definition of a labeled function?
 A: 
I hope you can see my point. Another examples: "Let $n\in \mathbb{N}$
  and define for all $x\in \mathbb{R}$ $f_{n}(x)=(n+1)x^{2}$ " Does this
  means $f:\mathbb{N}\times  \mathbb{R}\longrightarrow \mathbb{R}$?.

Yes. While it is sometimes convenient to think of the $f_1, f_2, ...$ as different functions, technically, the subscript is just another variable for the function.
$f_n (x) \equiv f(n,x)$
In your example, you could have written instead $f(n,x)=(n+1)x^{2}$
The subscript, as used here, is a typographical convention to make expressions easier to read. The subscript is usually a natural number as in your example. It is visually de-emphasized as a subscript.
EDIT

Yet another example:  "Let $X$ be a set. We talk about the identity
  function of $X$ to be defined as $I_{X}(y)=y$ for every $y\in X$".

This is less clear-cut. What set is $X$ an element of? There is no universal set of all sets, so the same analysis cannot apply. $I_X$ is just the identity function defined on the set $X$. The subscript is actually just part of the function name in this case. I don't think you would ever have to quantify over the $X$ in the $I_X$. Instead of writing $\forall X,I (P(I_X))$, for example, you could write $\forall X,I (\forall a\in X (I(a)=a) \to P(I))$
A: The subindex here has no formal definition; rather, it is used to illustrate that the specific function you are choosing is somehow associated with the element $p$.
This is a fairly common convention; if for each $x$ in some set you are choosing an object, rather than choosing a different letter for each one, you just subindex. In this way, it is clear that 
i) the different objects are all chosen in the same way (since they use the same letter) and 
ii) the object depends on the choice of $x$.
In a very real sense, in your second example, you could rewrite $f(x,n)$ instead of $f_n(x)$, and think of $f:\mathbb{R}\times\mathbb{N}\rightarrow\mathbb{R}$; you can also think of $f_{\circ}$ as a function from $\mathbb{N}$ to the set of mappings $\mathbb{R}\rightarrow\mathbb{R}$ (I tend to prefer to think of it in this sense).  However, in either case it is just a formalism for the same basic idea given above.
A: As has been noted by the other answerers, there is "no essential difference" between a mapping:
$$n \mapsto (f_n : \Bbb N \to\Bbb N)$$
and a mapping:
$$f: \Bbb N \times \Bbb N \to \Bbb N,(n,m) \mapsto f(n,m)$$
because we may intuitively put $f_n(m) = f(n,m)$ -- although only the first description is suitable for the use of the recursion theorem. Now let us investigate how precise this "no essential difference" claim can be made.
Let us denote things a bit more conveniently; we will write $\Bbb N^{\Bbb N}$ or $[\Bbb N \to \Bbb N]$ for the set of mappings $f: \Bbb N \to \Bbb N$.
Then the assignment $n \mapsto f_n$ can be viewed as a "function-valued function" $g: \Bbb N \to \Bbb N^{\Bbb N}, g(n) = f_n$ -- in place of $f_n(m)$, we thus (for the purpose of exposition) write the formal version $(g(n))(m)$ instead of $f_n(m)$. Because each $f_n$ is supposed to be a function $\Bbb N \to \Bbb N$, we can construct them using the recursion theorem (and this procedure is indeed intuitively connected to universal instantiation, because we construct it for fixed yet arbitrary $n$).
If we take a bit of a more abstract viewpoint, we will be able to discern what is actually going on. So let us consider $g: X \to Z^Y$ and $f: X \times Y \to Z$ instead; then the identification $(g(x))(y) = f(x,y)$ can be made explicit by:
\begin{align}
g:X \to Z^Y \quad&\longrightarrow\quad f(x,y) = (g(x))(y)\\
g(x) = f(x,-) \quad&\longleftarrow\quad f: X \times Y \to Z
\end{align}
where $f(x,-)$ is the function $f(x,-): Y \to Z, y \mapsto f(x,y)$.
For example, if $f(x,y) = y^x$, then $f(2,-)$ is the function $y \mapsto y^2$. Essentially, we treat $x$ as a constant when we consider $f(x,-)$.

At last, we get to the main (i.e., more abstract) point: This correspondence between $f$ and $g$ is actually very special, because it is independent of what $X,Y,Z$ are.
To make that claim more explicit, we introduce a new set $X'$ and a function $\alpha:X' \to X$. Furthermore, we define the following functions in terms of $\alpha$:
\begin{align}
\alpha_1: [X \to Z^Y] \to [X'\to Z^Y],& g \mapsto g \circ \alpha\\
\alpha_2: [X \times Y \to Z] \to [X' \times Y \to Z],& (\alpha_2 f)(x',y) = f(\alpha(x'),y)
\end{align}
(You are advised to play around with these definitions by plugging in $x',y,z$ at appropriate places, so as to get a feeling for what $\alpha_1$ and $\alpha_2$ do (it's really intuitive once you permeate the layer of abstraction).)
Now the upshot of this whole set-up is that if $f: X \times Y \to Z$ corresponds to $g: X \to Z^Y$ under the identification made above, then $\alpha_2(f):X' \times Y \to Z$ corresponds to $\alpha_1(g): X' \to Z^Y$. To give a bit of backing to the intuition, a similar approach can be given to prove that there is "no essential difference" between $(X\times Y)\times Z$ and $X \times (Y\times Z)$.
So in essence, the "way in which $f$ corresponds to $g$" "does not change" by changing $X'$ to $X$ via a map $\alpha$; that is, the correspondence does not affect any property that can be captured using functions (since there are similar statements for $Y$ and $Z$, it is nice to try and figure those out). So, this correspondence being "nice" in the above precise way, we often neglect that the difference exists altogether.
Now, should you be both intrigued and not yet exhausted, I advise you to check out Category Theory later in your mathematical life. It allows for making many kinds of "intuitive" statements about "natural correspondence" very precise.
A: I don't know what the description 1 means, but this description seems correct:
2) $f_p$  is a particular case of $f:N\times N \rightarrow N.$
What would be the problem with this description?
Edit 3: If the specific case where you want to apply the Recursion Theorem, I think you can proceed as follows. Take $s:N\rightarrow N$. For each $n\in N$, define $f(p,n)$ by $ f(p+1,n)=s\circ f(p,n)$ and $f(0,n)=n$ (supposing that $0\in N$ to simplify things). This $f(\cdot,n)$ is well defined by the Recursion Theorem. Now, since we defined it for each $n$, we have that $f:N\times N\rightarrow N$, as wanted.
Edit 2: Now in the example of the identity, this formalization doesn't seem correct. In that case, the Identity would be a function from sets and elements in that set to elements in that set. The problem is that we can't define a function (as far as I know) on "the set of all sets", because that's not a set. So there, the subindex is used naively. You can think of X as belonging to a particular class of sets (say subsets of $R$), or you may just think of it as not being an actual formal expression.
