As shown by Jeremy Rickard's answer, $S := S_0$ is usually not contained in an indecomposable direct summand. The purpose of this answer is to show the weaker statement

$S$ can be embedded into an indecomposable direct summand of $M$.

Proof: WLOG assume $S \neq 0$. Since $M$ is artinian, it is a direct sum of indecomposable submodules $M_1,..., M_m$. Let $n \le m$ be minimal such that there is an embedding (i.e. an injective hom. of modules) $S \hookrightarrow \oplus_{i=1}^n M_i$. If $n=1$ we are done. If $n > 1$ consider the composition $$S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i$$ If it's kernel is zero, $S$ embedds into $\oplus_{i=1}^{n-1}M_i$, in contradiction to the minimality of $n$. Hence, the kernel is non-zero and by simplicity of $S$, it's $S$, i.e. the composition is the zero map. Hence $$\text{im}(S \hookrightarrow \bigoplus_{i=1}^n M_i) \subseteq \ker(\bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i) = M_n$$ Thus the composition $S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow M_n$ is injective. QED.

As shown by Jeremy Rickard's answer, $S := S_0$ is usually not contained in an indecomposable direct summand. The purpose of this answer is to show the weaker statement

$S$ can be embedded into an indecomposable direct summand of $M$.

Proof: WLOG assume $S \neq 0$. Since $M$ is artinian, it is a direct sum of indecomposable submodules $M_1,..., M_m$. Let $n \le m$ be minimal such that there is an embedding (i.e. an injective hom. of modules) $S \hookrightarrow \oplus_{i=1}^n M_i$. If $n=1$ we are done. If $n > 1$ consider the composition $$S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i$$ If it's kernel is zero, $S$ embedds into $\oplus_{i=1}^{n-1}M_i$, in contradiction to the minimality of $n$. Hence, the kernel is non-zero and by simplicity of $S$, it's $S$, i.e. the composition is the zero map. Hence $$\text{im}(S \hookrightarrow \bigoplus_{i=1}^n M_i) \subseteq \ker(\bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i) = M_n$$ Thus the composition $S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow M_n$ is injective. QED.

Edit: Simpler proof: From $M=\oplus_{i=1}^m M_i$ we have $Hom(S,M)\cong \oplus_i Hom(S,M_i)$ and since the LHS is non-zero (it has the inclusion map), there is $i$ such that $Hom(S,M_i) \neq 0$. Let $0 \neq f: S \to M_i$ be a hom. By simplicity of $S$ we conclude $\ker f = 0$, i.e. $f$ is an embedding.