I'll just say my thoughts on some questions. Please verify them or suggest alternative solutions. Thanks in advance.
Let $R=\{(a_{ij}\in M_{3\times 3}(k))|\ a_{12}=a_{13}=0\}$ where $k$ is a field and $V=M_{3\times1}(k)$ as an $R-$module
(1) Find a composition series for $V$.
The set $V'=\{(0 **)^t \in M_{3\times 1}(k)\}$ is a submodule of $V$ which is simple and the quotient $V/V'\cong k$ is simple. Hence $0\subsetneq V'\subsetneq V$ is a composition series for $V$.
(2) Is $V$ a semisimple $R$-module?
No. If it was it should be a sum of simple $R-$modules but $V'$ is the only simple submodule of $V$ and $V\not=V'$.
(3) Is it true that every short exact sequence of $R-$modules $0\to A\to B\to C\to 0$ splits?
No. This is equivalent to saying that $R$ is a semisimple ring. If so then the $R-$module $V$ would be semisimple, contradiction.
Your answers are correct. The chain $0<V'<V$ contains all the submodules of $V$ and so is a composition series for $V$. You can show this by noting that if $\mathbf{v}=\begin{pmatrix}a&b&c\end{pmatrix}^t\in V$, then $\mathbf{v}$ generates $V$ if $a\ne 0$ and generates $V'$ if $a=0$ and one of $b,c$ is nonzero. It follows immediately that $V'$ has no complement and so $V$ is not semisimple, and $0\to V'\to V\to V/V'\to 0$ is a non-split short exact sequence.
As an alternative, you can answer (3) without (1) and (2) as follows. As you noted, if (3) has a positive answer, $R$ is semisimple. Let $I$ consist of all elements of $R$ that are $0$ everywhere but in the $2,1$ and $3,1$ entries. Then $I$ is a nonzero ideal with $I^2=0$. Thus $R$ cannot be semisimple.