= Solution
Write $e_{ij}$ for the <matrix units>. Left multiplication by $a=\begin{pmatrix}r&s\\0&t\end{pmatrix}$ gives
$$
ae_{11}=re_{11},\qquad ae_{12}=re_{12},\qquad ae_{22}=se_{12}+te_{22}.
$$
Thus a <composition series of a module> is
$$
\boxed{0\subset\mathbb Ce_{11}\subset\mathbb Ce_{11}\oplus\mathbb Ce_{12}\subset A.}
$$
Each quotient is one-dimensional, hence a <simple module>. The required <direct sum> is
$$
\boxed{{}_AA=Ae_{11}\oplus Ae_{22}
=\mathbb Ce_{11}\oplus(\mathbb Ce_{12}\oplus\mathbb Ce_{22}).}
$$
The first summand is simple. The second has a unique nonzero proper <submodule>, $\mathbb Ce_{12}$: any vector $ue_{12}+ve_{22}$ with $v\ne0$ generates $e_{12}$ by applying $e_{12}$ and then generates $e_{22}$. A decomposition into two nonzero summands would require a second one-dimensional <submodule>, which is impossible. Hence this summand is an <indecomposable module>.
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