For general groups $A, B$ with $B$ a subgroup of $A$, the quotient group $A/B$ is defined to be the elements of $A$ "modulo" the elements of $B$. In other words, the elements of $B$ all behave like a zero element in $A/B$.
So in $\Bbb Q / \Bbb Z$, which I'll call $G$ for simplicity, the integers behave like zero in $G$. So, for example, $\frac12$ and $\frac32$ are actually the "same" in $G$ because $\frac12 + 1 = \frac32$, and $1$ behaves like $0$ in $G$.
Technically speaking, the elements of $G$ (and quotient groups in general) are equivalence classes, and not simply numbers. This is why I put "same" in quotations above. While $\frac12$ and $\frac32$ are different numbers, they're part of the same equivalence class in $G$, because they differ by an integer in $\Bbb Q$ and all integers behave like zero in $G$, and so $\frac12$ and $\frac32$ "differ by zero" in $G$. Therefore for all intents and purposes we can treat them as being equal to each other in $G$. And we can denote the equivalence class by $[\frac12]$.
Why use $\frac12$ as our representative to denote the equivalence class? Why not $\frac32$? Or $\frac72$, or $-\frac92$? It's because it's best to choose the "simplest" representative.
Note that for every rational number $q$, we can find some integer $n$ such that $q-n \in [0,1)$. Therefore every rational number belongs to the same equivalence class as some rational number in $[0,1)$. And most would agree that the rationals in $[0,1)$ are the simplest to use as our equivalence class representatives.
Therefore $G = \Bbb Q/\Bbb Z$ consists of equivalence classes of the form $[q]$, where $q \in [0,1) \cap \Bbb Q$. And for each such $q$, the elements of $[q]$ look like $q+k$, where $k$ is any integer.