$$(\mathbb Z/6\mathbb Z)/(2\mathbb Z/6\mathbb Z)=\left\{ \left\{ 6\mathbb Z,2 + 6\mathbb Z, 4 + 6\mathbb Z\right\} ,\left\{ 1+6\mathbb Z,3 + 6\mathbb Z, 5 + 6\mathbb Z\right\} \right\} $$
Is that correct?
One proof for the third isomorphism theorem $(G/N)/(H/N)\stackrel{\sim}{\longrightarrow}(G/H)$ says we can view $H/N$ as a normal subgroup of $G/N$ - via the projection $\phi: G\to G/N$ and identifying $H/N$ as the image of $H$ under this projection (i.e., as $\phi(H)$). Can you give me an example where $H/N$ isn't equal to $\phi(H)$? In the example I've tried above they seem to be the exact same thing (both as sets and as groups).
As Peter says in the comments, if $\phi(x)=xN\in H/N$ for $x\in G$, then $\phi(H)=H/N$ by definition.
If we relax the strict definition of $\phi$ to allow it to be any homomorphism on $G$ with kernel $N$, then we can easily construct examples where $\phi(H)\neq H/N$. For example, if $G=\mathbb Z$ and $N=6\mathbb Z$, then we can form a map from $G$ to the subgroup of the complex numbers under multiplication generated by $e^{\pi i/3}$ via: $$ \phi(n)=e^{n\pi i/3} $$
The kernel of $\phi$ is then $6\mathbb Z$, and $\phi(H)$ is still isomorphic to $H/N$ for any subgroup $H\le G$, but they are not equal as sets. For example, if $H=2\mathbb Z$, then we have: \begin{align} H/N&=\{\mathbb Z,2+6\mathbb Z,4+\mathbb Z\}\\ \phi(N)&=\left\{1,e^{2\pi i/3},e^{4\pi i/3}\right\} \end{align}
The important thing to take away is that there is no group-theoretic distinction between these two situations. In each situation we have the same groups $G,N,H$, and we have a surjective homomorphism $\phi:G\to K$ for some group $K$ with $\ker\phi=N$ (so $K\cong G/N$, by the first isomorphism theorem). Then we get a correspondence of isomorphisms between the groups $H/N$ and the groups $\phi(H)$, where $H$ ranges over the subgroups of the group $G$. Whether $\phi(H)$ and $H/N$ are actually equal as sets is neither here nor there.