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.
Best Answer
Here is the argument written out in full so you can tell me which step you don't understand. Suppose $F_X$ is generated by a subset $Y$ with $|Y| < |X|$. This induces a surjection $F_Y \to F_X$. Abelianization gives a surjection $\text{Ab}(F_Y) \to \text{Ab}(F_X)$. Tensoring with $\mathbb{Q}$ gives a surjection from a vector space of dimension $|Y|$ to a vector space of dimension $|X|$; contradiction.