I want to prove that for a group $G$ and normal subgroups $H$ and $K$ where $H\cap K = \{e\}$ $$ G \simeq G/H\oplus G/K. $$
My thought is to use the map $g \mapsto (gH,gK)$. I can show that this is a homomorphism because $G/H$ and $G/K$ are groups. I can also show this is an isomorphism: This follows from $H\cap K = \{e\}$. So $gH = g'H$ and $gK = g'K$ implies that $g^{-1}g' \in H\cap K$.
But how can I show that the map is surjective?
On
If you assume (in addition to what you did) that $HK=G$, then, since $H$ is normal and $H\cap K$ is trivial, $G=H\rtimes K$. What is left is to show that the action of $K$ (by conjugation) on $H$ is trivial.
Then you only need to show that $khk^{-1}h^{-1}=e$ (for all appropriate $k,h$). For that, you need to use normality of $K$ and $H$.
This will imply that $G=K\oplus H$. But then obviously $G/H\cong K$ and $G/K\cong H$
It certainly is injective, since the kernel is $H\cap K$. But it is not surjective without other hypotheses. For instance, in an abelian group, all subgroups are normal, but $(\mathbf Z/p\mathbf Z\times\mathbf Z/q\mathbf Z)\times\mathbf Z/r\mathbf Z$ is not isomorphic to $\mathbf Z/p\mathbf Z\times\mathbf Z/q\mathbf Z$ since the former has $pqr$ elements and the latter $pq$.
Another counter example is this: any two subgroups $H,K$ of a group $G$ with distinct prime orders $p$ and $q$ intersect trivially. There does not result that $G\simeq H\times K$: otherwise, it would mean that any group with two subgroups of prime orders $p$ and $q$ has order $pq$, which is absurd (what if $G$ has $3$ subgroups of distinct prime orders?)