Let $N\vartriangleleft G$ be a normal subgroup in $G$. The homomorphism theorem states:
If $f\colon G\longrightarrow H$ is a homomorphism of groups and $N$ is given as above, where also $N\subset \ker f$, then there is a unique homomorphism $\overline f\colon G/N\rightarrow H$ such that $\overline f(aN)=f(a), \ \forall a\in G$.
We need that $N$ is a normal subgroup of $G$, but why?
Consider another subgroup $L$ of $G$ and define $G/L$ with the product
$(aN,bN)\mapsto (ab)N$ for its corresponding cosets. (lets consider the left ones here)
This would be a group, too, would it?
I simply don't see, why the group has to be normal in order for the homomorphism theorem to work. It must have something to do with the group structure on $G$.
If the subgroup is not normal you can't define $G/L$ as a group.
When you try to define the operation on $G/L$ by $(aL) \cdot (bL) = (abL)$ if $L$ is not normal the operation isn't well defined, in the sense that the product depends on $a$ and $b$, and not only on the class $(aL)$ and $(bL)$. You could find $a_0 \in (aL)$ and $b_0 \in (bL)$ such that $(a_0 L) \cdot (b_0L) = (a_0 b_0L)$ but $(a_0 b_0 L) \neq (abL)$.