I know that the Jacobson radical of a module $M$ is the intersection of all submodules $N$ of $M$ such that $M/N$ is simple. However, why there is another definition saying that the Jacobson radical of a module $M$ is the intersection of all submodules $N$ of $M$ such that $M/N$ is semisimple?
How are these two definitions equivalent?
For a sake of simplicity of notation, let $A= \{ N \le M : M/N \mbox{ is simple} \}$ and $B= \{ N \le M : M/N \mbox{ is semisimple} \}$. Obviously $A \subseteq B$. Let's prove by double inclusion that $$\bigcap_{N \in A}N = \bigcap_{N \in B}N$$ The inclusion "$\supseteq$" is a trivial consequence of $A \subseteq B$.
Conversely, for all $x \in \left( \bigcap_{N \in B}N \right)^c$ there exists some $N \in B$ such that $x \notin N$. Denote by $p: M \to M/N$ the canonical projection. We know that every semisimple module is a direct sum of simple modules, so write $$M/N = \bigoplus_{i \in I} S_i$$ where $\{ S_i \}_{i \in I}$ is some family of simple modules indexed by a set $I$.
For all $i \in I$ consider the canonical projection $\pi_i : M/N \to S_i$ and the composition $$\pi_i \circ p : M \longrightarrow S_i$$ Obviously $K_i=\ker (\pi_i \circ p)$ is a submodule of $M$ such that $M/K_i \cong S_i$ is simple: in other words $K_i \in A$. You can check that $$N=\ker p = \bigcap_{i \in I} K_i$$ Since $x \notin N$, there exists some $K_i$ not containing $x$. So that $$x \notin \bigcap_{N \in A}N$$
This proves the opposite inclusion.