In $R$-Mod Category, example for $B\cong A \oplus C \nRightarrow 0 \to A \to B \to C \to0$ splits. From splitting lemma, we know in $R$-Mod Category, short exact sequence 
$0 \to A \stackrel{f}{\rightarrow} B \stackrel{g}{\rightarrow} C \to0$ splits 
if it satisfies one of the following equivalent conditions:
$(1)\ \exists f_1\in\text{Hom}(B,A)\text{ s.t. } f_1\circ f=\text{Id}_A$.
$(2)\ \exists g_1\in\text{Hom}(C,B)\text{ s.t. } g\circ g_1=\text{Id}_C$.
$(3)\ \text{Im }f=\text{Ker }g$ is direct summand of $B$.
$(4)\ \exists \text{ isomorphism } h:B\to A \oplus C \text{ s.t. } $
$h \circ f \text{ is natural injection and }g \circ h^{-1} \text{ is natural projection.}$
And $0 \to A \stackrel{f}{\rightarrow} B \stackrel{g}{\rightarrow} C \to0$ splits $\implies B\cong A \oplus C$.
If we only have $B\cong A \oplus C$, is there any example for
$0 \to A \to B \to C \to0$ doesn't split?

Related questions:
$(1)$ Example for infinitely generated modules.
$(2)$ Example for abelian groups.
As proved in answer below, it's ture for modules on commutative ring with finite length.

Special thanks for jgon, for his knowledge, time, patience and friendliness.
 A: Let $R$ be a commutative ring and let 
$$\begin{equation}\label{eq:ses}0\to A\to B\to C\to 0\tag{*}\end{equation}$$
be a short exact sequence of modules of finite length over $R$. We show that if $B\cong A\oplus C$, then $\eqref{eq:ses}$ splits.
Since $B\cong A\oplus C$, we have that $$\newcommand\Hom{\operatorname{Hom}}\Hom_R(C,B)\cong\Hom_R(C,A)\oplus\Hom_R(C,C).$$ 
Hence $$\ell(\Hom_R(C,B))=\ell(\Hom_R(C,A))+\ell(\Hom_R(C,C))$$ where $\ell$ denotes length.
Write $f:B\to C$ for the map in $\eqref{eq:ses}$. Apply $\Hom_R(C,-)$ to the sequence $\eqref{eq:ses}$. 
From the left exactness of $\Hom_R(C,-)$, we obtain the exact sequence
$$0\to \Hom_R(C,A)\to \Hom_R(C,B)\to \Hom_R(C,C),$$
where the map $f_* : \Hom_R(C,B)\to \Hom_R(C,C)$ is given by $g\mapsto f\circ g$. 
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that 
$$0\to \Hom_R(C,A)\to \Hom_R(C,B)\to \Hom_R(C,C)\to N \to 0$$
is exact, and hence that $\ell(N)=0$. 
Hence $f_*$ is surjective, which shows that the sequence $\eqref{eq:ses}$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
