Assume $H,G$ are simple groups. Can we prove that either $H\unlhd HG$ or $H\unlhd GH$?

If $H_1$ and $H_2$ are simple subgroups of a larger group $G$, and $H_1 H_2$ is a subgroup of $G$ it is not true in general that $H_1$ and $H_2$ are both normal in $H_1H_2$. An example might be

$$G = \{ \begin{pmatrix} x & y \\ & 1 \end{pmatrix} : 0 \neq x \in \mathbb F_3, y \in \mathbb F_3\}$$

$$H_1 = \{ \begin{pmatrix} x \\ & 1 \end{pmatrix} 0 \neq x \in \mathbb F_3\}$$

$$H_2 = \{ \begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} y \in \mathbb F_3\}$$

Here $\mathbb F_3$ is the field with three elements, $H_1 \cong \mathbb Z/2\mathbb Z$ and $H_2 \cong \mathbb Z/3\mathbb Z$ are both simple groups, and $H_1H_2 = G$ is a group, yet $H_1$ is not normal in $H_1H_2$.

But in your case, of course $H_1$ is normal in $H_1 \times H_2$, $H_1 \times H_2$ is normal in $H_1 \times H_2 \times H_3$, and so on. This gives as you say a composition series whose quotients are the simple groups $H_1,H_2, ...$. You can use Jordan-Holder to argue that $n=m$ and that with some reordering, each $H_i \cong G_i$.

$$\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} \begin{pmatrix} x \\ & 1 \end{pmatrix} \begin{pmatrix} 1 & -y \\ & 1 \end{pmatrix}$$

$$\begin{pmatrix} x & y \\ & 1 \end{pmatrix}\begin{pmatrix} 1 & -y \\ & 1 \end{pmatrix} = \begin{pmatrix} x & y-xy \\ & 1 \end{pmatrix}$$

No. Any transposition of $S_3$ generates a simple subgroup that is not normal in $S_3$, even though $S_3$ is the product of the subgroup generated by any transposition and the subgroup generated by any $3$-cycle.