A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$

There is no counterexample with $A,B,C$ finitely generated abelian groups. There is, more generally, no counterexample with $A,B,C$ finitely generated modules over any noetherian ring $R$.

To see this, consider the exact sequence

$$0\rightarrow\mathrm{Hom}(C,A)\rightarrow\mathrm{Hom}(B,A)\rightarrow \mathrm{Hom}(A,A).$$

The original sequence splits if and only if this sequence is exact on the right. If $A$, $B$ and $C$ are of finite length as modules, this follows immediately just by counting lengths. Otherwise, it's enough to show exactness after localizing and completing at an arbitrary prime $P$, and for this it's enough to show exactness after tensoring with $R/P^n$, and for this you can assume the lengths are finite, which is the case we've already dealt with.

I'm sorry, but my answer uses Ext groups, which may not be in the scope of your course.

I don't know of a simple example with finitely generated abelian groups for the following reason: if $0 \rightarrow \mathbb Z \rightarrow E \rightarrow \mathbb Z / n \mathbb Z \rightarrow 0$ is an extension represented by $r+n \mathbb Z \in \operatorname{Ext}^1(\mathbb Z,\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$, then one can show that $E \cong \mathbb Z \oplus \mathbb Z/d\mathbb Z$ where $d$ is the highest common factor of $r$ and $n$. If you assume this sequence doesn't split, then $r$ is not a multiple of $n$ and $d<n$, so the middle term will never be isomorphic to the direct sum of the of the first and third.

However, you can get a simple example using modules in the following way. Let $G=\langle g\rangle$ be an infinite cyclic group and let $\mathbb Z G$ be the associated commutative group ring. Consider $0 \rightarrow \mathbb Z \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0$ a short exact sequence of $\mathbb Z G$-modules where the first map is inclusion into the first component and the second map is projection onto the second. Let $g$ act on $\mathbb Z \oplus \mathbb Z$ by $\bigl(\begin{smallmatrix} 1&1\\ 0&1 \end{smallmatrix} \bigr)$. Then $\operatorname{Ext}^1_{\mathbb Z G}(\mathbb Z, \mathbb Z) \cong \mathbb Z$ and this extension corresponds to $1$. If you let $g$ act by $\bigl(\begin{smallmatrix} 1&n\\ 0&1 \end{smallmatrix} \bigr)$ it corresponds to $n$. As long as $n>0$ this is a non-split short exact sequence where the direct sum of the outside terms are isomorphic to the middle one. However, this example is far from elementary, and I apologize for that.

See the answer in this question:

In R-Mod Category, example for $B\cong A \oplus C \nRightarrow 0 \to A \to B \to C \to0$ splits.

It's similar to WillO's.