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.