No, $N$ does not contain all rational numbers in $[0,1)$. In fact, it contains one and only one rational number from $[0,1)$ (though we don't know which, since $N$ is defined non-constructively via the Axiom of Choice).
As is mentioned, $N$ contains exactly one member from each equivalence class under $\sim$. That means that:
(i) For every $x\in[0,1)$, there exists $y\in N$ such that $x\sim y$, i.e., such that $x-y\in\mathbb{Q}$; and
(ii) For every $x,y\in N$, if $x\sim y$ (that is, if $x-y\in\mathbb{Q}$), then $x=y$.
In particular, we know that there exists $x\in N$ such that $0\sim x$; since this requires that $-x\in\mathbb{Q}$, that means that $x$ is rational. But if $y\in N$ is rational, then $x-y\in\mathbb{Q}$, hence $x\sim y$, so by (ii) we conclude that $x=y$. So $N$ contains one and only one rational from $[0,1)$.
The motivation fro the definition of $N_r$ is that it is a "wrapped around" translate of $N$ by $r$. That is, take every number in $N$, add $r$ to it; if it is still in $[0,1)$, keep it there; otherwise, shift it back by $1$ (so that it falls inside $[0,1)$). These are just rational translates of $N$, adjusted to make sure they stay in the interval.
The reason you want the $N_r$ is that you will show that these countably many translated copies of $N$ will "cover" a nontrivial interval; they are pairwise disjoint (that is, if $r\neq s$, then $N_r\cap N_s=\emptyset$), so the fact that countably many disjoint sets cover a nontrivial interval will imply that, if they are all measurable, then their measure must be positive (by $\sigma$-additivity). Since they are all just translates of the same set $N$, they are each measurable if and only if $N$ itself is measurable, and they will all have the same measure. But then measurability of $N$ would imply that the union of the countably many disjoint translated copies of $N$ will have infinite measure, which will also be impossible (since they are all contained in $[0,1)$, so by monotonicity, their measure, if it exists, will be finite). These things put together will imply that $N$ cannot possibly be measurable.
(As to how one would come up with it in the first place, you'll have to ask Vitali...)