$G/N$ read as $G$ modulo $N$.

Your intuition is a bit off but not entirely. This quotient group essentially creates a group in which all of $N$ is now the identity (I'm sure you know we can only do this if $N\triangleleft G$). So, for example, let's consider the integers along with the normal subgroup $6\mathbb{Z}$. Then we can define a quotient group

$$\mathbb{Z}/6\mathbb{Z}=\{6\mathbb{Z},1+6\mathbb{Z},2+6\mathbb{Z},3+6\mathbb{Z},4+6\mathbb{Z},5+6\mathbb{Z}\}$$

We can add these together in ways such as $$(3+6\mathbb{Z})+(4+6\mathbb{Z})=1+6\mathbb{Z}$$

If you've been dealing with the groups of integers under modular addition you'll notice this quotient group we've defined does precisely the same that we would expect of $\mathbb{Z}_6$. And, indeed, the two groups are isomorphic.

One important use of this idea as well is called the First Isomorphism Theorem. Let's say we've set up a homomorphism that's not injective (i.e. a nontrivial kernel). Well, the kernel is always a normal subgroup, so we can "mod out" by the kernel, essentially condensing our entire kernel into one element. We then have something that's injective. This is stated more formally as:

Let $\phi:G\rightarrow H$ be a group homomorphism. Then,

$$G/\ker\phi\cong\text{im}\phi$$

This is extremely useful in proving certain isomorphisms.

A little more generally, the use of the word "modulo" is connected to equivalence relations. Given a set $X$ and an equivalence relation $R$ defined on $X$, the set of equivalence classes is frequently denoted $X/R$ and called "X mod R".

In your example, the subgroup $N$ can be used to define an equivalence relation on $G$: Declare elements $g, h \in N$ to be equivalent if $g^{-1}h \in N$. The equivalence classes are precisely the left cosets of $N$ in $G$. There's thus a very nice affinity between the general notation $X/R$ and the group-specific notation $G/N$.