Q:

Intuition: groups, quotient groups, cosets, homomorphisms. 

If we start with the group of rational numbers $\mathbb{Q}$
and the subgroup of $\mathbb{Q}$; $\mathbb{Z}$ the integers,
and then form the quotient group $\mathbb{Q}$/$\mathbb{Z}$
we have that this quotient group consists
of all the cosets of $\mathbb{Z}$ in $\mathbb{Q}$.
These are of the form $a + \mathbb{Z}$, $a \in \mathbb{Q}$.
E.g. $...\frac{2}{3}, \frac{5}{3}, \frac{8}{3}...$ 
i.e. the coset $\frac{2}{3} + \mathbb{Z}$.
These cosets are the elements of the quotient group $\mathbb{Q}$/$\mathbb{Z}$ and is a partiton of $\mathbb{Q}$.
Then we have that $\mathbb{Q}$/$\mathbb{Z}$
is homomorphic to the circle $S^1$.
$\mathbb{Q}$/$\mathbb{Z} \:\cong \:S^1$
This is because the cosets 
are parameterized by elements belonging to
the interval $[0,1]$. Every coset has 
exactly one element in it (as long as
we make the identification that $0 = 1$, because
it belongs to the original subgroup $\mathbb{Z}$)
$\mathbb{Q}$/$\mathbb{Z}\: \cong \:S^1$ ($\cong [0,1]$ with $0=1$)
I think these things are rather difficult and
i wonder if my intuition, understanding,
of them and the connection between them are ok, or
if there are some flaws? 

Clarification from the discussion in the comments:
The statement $\mathbb{Q}$/$\mathbb{Z}\: \cong \:S^1$
comes from Wildberger's video lectures and what he probably 
means is that these are homeomorphic as spaces. 
But even this is wrong.  
Conclusion: Do not watch his videos :)

A:

As stated in the comments $\mathbb Q/\mathbb Z$ is not isomorphic to $\mathbb S^1$ which is the unit circle.  What we have is the following isomorphism with the multiplicative group of roots of unit 
$$\begin{align}\varphi: \mathbb Q/\mathbb Z &\to C^{\times}\\\frac{p}{q} + \mathbb Z &\mapsto e^{\frac{2\pi i p}{q}}\end{align}$$

