Let's work out explicitly what the quotient looks like by a subgroup of order 4. Specifically, let's quotient by $\left = \{1,A,A^2,A^3\}$.
$Q/\left$ is, by definition, the set of cosets of $\left$ in $Q$. As pointed out, there must be two cosets because 8/4=2. One of these cosets, $S_1$, contains the elements of $\left$, and the other, $S_2$, contains the leftovers: $\{B,AB,A^2B,A^3B\}$.
Strictly speaking, these two cosets are the elements of $Q/\left$. The group operation is defined as the usual product for two subsets of a group. If $S$ and $T$ are subsets of group $G$, their product $ST$ is said to be the set $\{st|s\in S, t\in T\}$.
Under this product, the two cosets of $\left$ in $Q$ form a group. You can verify directly that $S_1S_1 = S_2S_2 = S_1$ and $S_2S_1=S_1S_2 = S_2$. These relations demonstrate that $Q/\left$ is cyclic of order 2.
This is a relatively unexciting example. If there were more than two cosets, you might have to compute several pairwise products before discovering the structure of the quotient group. Because $|Q/\left|=2$, there was only one possibility for a group structure. Also, if you were to take $G/H$ with $H$ not normal in $G$, you would find that the cosets would not form a group under the product we defined.