The set of all classes mod $R$ is denoted : $\frac{G}{R}$, with $(G, \times)$ a group and $R$ an equivalence relation on $G \times G$
For example : $\mathbb{Z} / n\mathbb{Z} = \{\bar{0}, ..., \overline{n-1} \}$
My question is then : Why the set of all classes mod $R$ is denoted $\frac{G}{R}$ ? I mean I don't see the link with the division notation ?
Here's is a tentative of explanation.
If you think a minute about the meaning of taking quotients modulo equivalence relations, you will see that you are just partitioning the original set into subsets of equivalent elements. In some sense, this is like "dividing a set in many parts", with the caveat that they are not all the same.
In general, it is a good idea to have in mind the picture "quotient" $\leftrightarrow$ "partition", which indeed corresponds to reality. Take for instance $G=\mathbf{Z}$ the additive group of integers and let $R$ the relation you mention: $x\,R\,y$ if and only if $x-y$ is divisible by $n$.
Every equivalence class of an element $x\in \mathbf{Z}$ is made up of the following elements: $$[x]=\{y\in \mathbf{Z}:y-x=nz\text{ for some }z\in \mathbf{Z}\}=\{x+nz:z\in \mathbf{Z}\}$$
It is immediate to see that these equivalence classes form a partition of $\mathbf{Z}$, in the sense that their union is the whole group of integers and they are two-by-two disjoint.
As a side note, don't confuse the quotient set (the family of all equivalence classes, namely a partition of the original set) with a set of representatives for it: in your example you say that $\mathbf{Z}/n\mathbf{Z}=\{[0],\ldots,[n-1]\}$ but this is only a set of representatives for the quotient set; you could have chosen any set $T\subseteq \mathbf{Z}$ such that $\mathrm{card}(T\cap [x])=1$ without difference.