Consider the ring $$ R = \begin{pmatrix} \mathbb{Q} & 0 \\ \mathbb{R} & \mathbb{R} \end{pmatrix} $$ as a left $R$-module over itself. I want to show that the following is a composition series $$ 0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\subset \begin{pmatrix} 0 & 0 \\ \mathbb{R} & 0 \end{pmatrix}\subset \begin{pmatrix} 0 & 0 \\ \mathbb{R} & \mathbb{R} \end{pmatrix}\subset \begin{pmatrix} \mathbb{Q} & 0 \\ \mathbb{R} & \mathbb{R} \end{pmatrix}=R $$ for $R$.
So I consider the quotients of the terms, for example $ \begin{pmatrix} \mathbb{Q} & 0 \\ \mathbb{R} & \mathbb{R} \end{pmatrix} / \begin{pmatrix} 0 & 0 \\ \mathbb{R} & \mathbb{R} \end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?
Similarly how can I do the same thing for the quotient $\begin{pmatrix} 0 & 0 \\ \mathbb{R} & 0 \end{pmatrix} / \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ ?
That alone is probably not a transparent enough explanation of why it is simple. You can, if you like, start with the fact it is isomorphic to $\begin{pmatrix} \mathbb Q & 0 \\ 0 & 0 \end{pmatrix}$ as a left $R$ module, and note that for any two nonzero elements $\begin{pmatrix}x & 0\\0&0\end{pmatrix}$ and $\begin{pmatrix}y & 0\\0&0\end{pmatrix}$, you can find $\begin{pmatrix}q & 0\\r&s\end{pmatrix}$ such that $\begin{pmatrix}q & 0\\r&s\end{pmatrix}\begin{pmatrix}x & 0\\0&0\end{pmatrix}=\begin{pmatrix}y & 0\\0&0\end{pmatrix}$. (When a ring acts transitively on the nonzero elements of a module, that means it is simple.)
Similar logic holds for $\begin{pmatrix} 0 & 0 \\ \mathbb{R} & \mathbb R \end{pmatrix} / \begin{pmatrix} 0 & 0 \\ \mathbb R & 0 \end{pmatrix}\cong\begin{pmatrix} 0 & 0 \\ 0&\mathbb R \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ \mathbb{R} & 0 \end{pmatrix} / \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\cong \begin{pmatrix} 0 & 0 \\ \mathbb R & 0 \end{pmatrix}$