Let $N_1, N_2$ be two normal subgroups of the group $G$, with $N_1\times N_2$ denoting the external direct product. We need to show that $G$ is an internal direct Product of $N_1, N_2$ iff $$ \varphi:N_1\times N_2 \rightarrow G, \, \, (a_1,a_2) \mapsto a_1 \cdot a_2 $$ is an isomorphism.
Showing that assuming the G is an internal product then $\varphi$ is an isomorphism is pretty straight forward.
If $\varphi$ is an isomorphism obviously $N_1 \cdot N_2 =G$. I have troubles understanding $$ \varphi(N_1\times \{e\})=N_1 \, \, ,\varphi(\{e\}\times N_2)=N_2 \Rightarrow N_1 \cap N_2=\varphi(e,e)=e $$
The only idea I could come up with was that $N_1 \cap N_2 = \varphi(N_1\times \{e\}) \cap \varphi(\{e\}\times N_2)= N_1 e \cap e N_2 = ee = \varphi(e,e)=e$
I would appreciate your help.
If I understand correctly, you are struggling to show that $N_{1}$ and $N_{2}$ have trivial intersection (just the identity). To see this, suppose that there exists some element $a$ that lies in both sets. The pairs $(a, 1)$ and $(1, a)$ both get mapped to $a$ by the isomorphism you have defined. Since an isomorphism is injective by definition, $a = 1$. This completes the proof.