Yes, $N$ is the subgroup generated by all squares. $N$ is normal in $F$, because if $w^2$ is an element of the generating set for $N$, and $x\in F$, then $xw^2x^{-1} = (xwx^{-1})^2$ is also an element of $N$. So for every $x\in F$ we have $xNx^{-1}\subseteq N$, hence $N$ is normal.
When $p$ is a prime, an abelian $p$-group is simply an abelian group all of whose elements have order a power of $p$ (the group could be finite or infinite). An elementary abelian $p$-group is an abelian $p$-group in which every element satisfies $a^p = 1$ (and so, every element except for the identity is of order exactly $p$; this is with multiplicative notation, you would have $pa=0$ if you are using additive notation for your group). So the assertion is that $F/N$ is abelian, and the square of every element is the identity.
The fact that the square of every element of $F/N$ is the identity follows because every square is in $N$: if $fN\in F/N$, then $(fN)^2 = f^2N = 1N$. And the fact that $F/N$ is abelian now follows from the well-trod fact that a group in which the square of every element is the identity must be abelian (since $1 = (ab)^2 = a^2b^2$, so $abab=aabb$, hence $ba=ab$ by cancellation).
An elementary abelian $p$-group is always a vector space over $\mathbb{F}_p$, the field with $p$-elements: given $\alpha\in\mathbb{F}_p$, let $a\in\mathbb{Z}$ by any integer mapping to $\alpha$. Then, assuming your group is written additively, define $\alpha\cdot g$ as $\alpha\cdot g= ag$. Since $pg=0$, this is well defined and makes the abelian group into a vector space.
So here, you have that $F/N$ is an abelian $2$-group, and therefore is a vector space over the field of $2$-elements.
In fact, the images of the free generating set $X$ in $F/N$ form a basis for this vector space: since $X$ spans $F$, its images span $F/N$. And if you have a nontrivial linear combination between them, then it must be of the form
$$\overline{x_1}+\cdots + \overline{x_n} = \mathbf{0}$$
where $\overline{g}$ is the image of $g\in F$ in the quotient, and $x_1,\ldots,x_n$ are pairwise distinct elements of $X$. But this means that $x_1\cdots x_n\in N$, that is, that it is the square of an element of $F$, and this is easily shown to be impossible.
So $F/N$ is a vector space over $\mathbb{F}_2$, and has a basis of cardinality $|X|$. How many elements does a vector space of dimension $\kappa$ over $\mathbb{F}_2$ have? If $\kappa$ is finite, then it has $2^{\kappa}$ elements. If $\kappa$ is infinite, then it has $\kappa$ elements. So if $X$ is infinite, then $|X|=|F/N|$.
Now suppose that $F_X$ and $F_Y$ are isomorphic. Then the isomorphism maps $N=\langle w^2\mid w\in F_X\rangle$ to $M=\langle z^2\mid z\in F_Y\rangle$, so we get that $F_X/N\cong F_Y/M$. If one of them is finite, then they both are; if one of them is infinite, then they both are. If both are finite, then the cardinality of $F_X/N$ is $2^{|X|}$, and the cardinality of $F_Y/M$ is $2^{|Y|}$, and since they are isomorphic groups, then $|X|=|Y|$. If they are both infinite, then $F_X/N$ has cardinality $|X|$, and $F_Y/M$ has cardinality $|Y|$, and since they are isomorphic their cardinalities are the same, so $|X|=|Y|$ as well. This proves the result.
There is a simpler way: the result holds for abelian free groups (tensor up to $\mathbb{Q}$ to reduce to the vector space $K$), and then show that $F_X^{\mathrm{ab}}$ is the free abelian group on $X$.
To answer final question: "word length" depends on the free basis. There is always a choice of basis for $F_Y$ that makes $\varphi$ preserve word length (simply take the basis $\varphi(X)$), but in general it need not. Take $X=\{x_1,x_2\}$, $Y=\{y_1,y_2\}$, and first map $F_X$ to $F_Y$ the obvious way ($x_1\mapsto y_1$, $x_2\mapsto y_2$), and then compose with a suitable inner automorphism of $F_Y$. For example, composing with conjugation by $y_1$ maps $x_1\mapsto y_1$ and $x_2\mapsto y_1y_2y_1^{-1}$. Composing with conjugation by $y_1y_2y_1^{-1}$ makes it even worse:
$$\begin{align*}
x_1 &\mapsto (y_1y_2y_1^{-1})y_1(y_1y_2^{-1}y_1^{-1}) = y_1y_2y_1y_2^{-1}y_1^{-1},\\
x_2 &\mapsto (y_1y_2y_1^{-1})y_2(y_1y_2^{-1}y_1^{-1}) = y_1y_2y_1^{-1}y_2y_1y_2^{-1}y_1^{-1},
\end{align*}$$
mapping the two generators to words of length 5 and 7, respectively; of course, you can make it pretty much as bad as you want using this idea.
It does not follow that for groups which are not finitely generated the notion of hyperbolicity makes sense.
What follows, instead, is that for any pair $(G,S)$ such that $G$ is a group and $S$ is a generating set, the notion of hyperbolicity is well-defined: the Cayley graph $\Gamma(G,S)$ is certainly defined (and it is connected), and hyperbolicity of $\Gamma(G,S)$ certainly makes sense.
The trouble is that without any control on the cardinality of $S$, it is possible to have a group $G$ and two generating sets $S_1,S_2$ such that $\Gamma(G,S_1)$ is hyperbolic whereas $\Gamma(G,S_2)$ is not.
In fact, here's an example: $G = \mathbb Z^2$ with $S_1 = \{(\pm 1,0),(0,\pm 1)\}$ and with $S_2 = \mathbb Z^2$. The Cayley graph $\Gamma(G,S_1)$ is not hyperbolic whereas $\Gamma(G,S_2)$ is bounded and is therefore hyperbolic. That's an example where $G$ is in fact finitely generated, but one can easily cook up examples where $G$ is not finitely generated, e.g. $G = \mathbb Z^\infty$ with the "standard" generating set $S_1 = \{\pm e_i \mid i \ge 1\}$ is not hyperbolic, but with the generating set $S_2=G$ it is hyperbolic.
What happens in the finitely generated case is that if $S_1,S_2$ are two finite generating sets for the same group $G$, then $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are quasi-isometric to each other, and hyperbolicity is a quasi-isometry invariant, so $\Gamma(G,S_1)$ is hyperbolic if and only if $\Gamma(G,S_2)$ is hypebolic. Thus one says that a finitely generated group $G$ is hyperbolic if and only if some (hence any) finite generating set $S$ results in a hyperbolic Cayley graph $(G,S)$.
The summary statement is that hyperbolicity of a finitely generated group is well-defined independent of the choice of finite generating set, but take away those bold faced words and the statement becomes false.
ADDED: In answer to a followup question in the comments, every Gromov hyperbolic space (i.e. every geodesic metric space satisfying the thin triangle property) has a Gromov boundary.
And, given a group $G$ and two generating sets $S_1$, $S_2$, if $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are both hyperbolic, then both of them do have a Gromov boundary. And if $S_1,S_2$ are both finite then one can use the quasi-isometry between $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ to deduce that their Gromov boundaries are homeomorphic; so in the finitely generated case you are free to speak about THE (homeomorphism class of the) Gromov boundary.
However, without the assumption that $S_1,S_2$ are both finite, it need not be true that the Gromov of $\Gamma(G,S_1)$ is homeomorphic to the Gromov boundary of $\Gamma(G,S_2)$.
For a counterexample, take your favorite infinite, finitely generated, hyperbolic group $G$. With any finite generating set $S_1$, the Gromov boundary is nonempty. With the infinite generating set $S_2=G$, the Gromov boundary is empty.
Anyway, it follows that to speak of THE Gromov boundary of $G$ is nonsensical in the infinitely generated case.
Best Answer
The equality $s=b$ holds for all finitely-generated infinite groups, and this does not relate to hyperbolic groups or even to exponential growth. It is a straighforward consequence of the isoperimetric inequality, by which for every finitely-generated infinite group $G$, for every $n$ one has $$\left|B_{n}\right|\leq\left(n+1\right)\left|S_{n}\right|.$$ In fact, for non-amenable groups and in particular non-elementary hyperbolic groups, one can replace $\left(n+1\right)$ in the above inequality by a positive constant depending only on the group.
For a proof of this see for instance Chapter 12.2 and in particular Theorem 12.14 in [2].
[2] Mann, Avinoam, How groups grow., London Mathematical Society Lecture Note Series 395. Cambridge: Cambridge University Press (ISBN 978-1-107-65750-2/pbk). ix, 199 p. (2012). ZBL1253.20032.