From Printers book.Ex. H2 chapter 14
Theres a hint ($hk=kh \iff hkh^{-1}k^{-1}=e$), and using the fact H,K are normal. I did it using the hint and it was ok.
My question is why cant I just say that by definition of H or K being normal $aH=Ha$, for any $a\in g$. So since $k\in G, kH=Hk$ i.e. $kh=hk , \forall h \in H$
Or do I have a flawed understanding of definition of normal subgroup?
Your proposed argument only uses the normality of $H$; it does not use the normality of $K$, nor does it use the fact that $H\cap K=\{e\}$. If your argument were correct, it would imply that every normal subgroup in every group is central... which is patently false.
The error is that the equality $kH=Hk$ holds at the level of sets. It says that $$\{ kx\mid x\in H \} = \{yk\mid y\in H\}$$ (I'm using different variables to highlight the problem). That just says that for each $x\in H$ there exists a $y\in H$ such that $kx=yk$, and for each $y'\in H$ there exists an $x'\in H$ such that $y'k = kx'$. But it does not say that $x$ and $y$ are equal, or that $y'$ and $x'$ are equal. So you cannot conclude that $kx=xk$ for all $x\in H$, $k\in K$.
For an example where $H\triangleleft G$, $K\triangleleft G$, but not every element of $H$ commutes with every element of $K$, take $G=Q_8$, the quaternion group of order $8$, $H=\langle i\rangle$, $K=\langle j\rangle$. Note that $ij\neq ji$. And you can see that, for example, $$\begin{align*} jH &= \{j1, j(-1), ji, j(-i)\} = \{j, -j, -k, k\}\\ Hj &= \{1j, (-1)j, ij, (-i)j\} = \{j,-j, k, -k\} \end{align*}$$ so that $jH=Hj$, but that's because $ji=(-i)j$ and $j(-i)=ij$, not because $jh=hj$ for all $h\in H$.