A chain complex $A_{*}$ is contractible if and only if $A_{*}$ is acyclic and $\iota: Z_{n}A \hookrightarrow A_{n}$ is a split monomorphism. I already proved the forward direction. But I'm stucked with the backward direction. Here's my proof attempt:
Assume that $A_{*}$ is acyclic and $\iota: Z_{n}A \hookrightarrow A_{n}$ is a split monomorphism. Then $H_{*}A = 0_{*}$, and so $Z_{n}A = B_{n}A$, and there exists $j: A_{n} \to Z_{n}A$ such that $j \circ \iota = 1_{Z_{n}A}$. To prove that $A_{*}$ is contractible, I must show that there exists $h : A_{n} \rightarrow A_{n+1}$ such that $h : g \circ f \simeq 1_{A_{*}}$, where $f : A_{*} \rightarrow 0_{*}$ and $g : 0_{*} \rightarrow A_{*}$.
What I have deduced so far is that if such $h$ exists, then $dh + hd = 1_{A_{*}}$. I tried looking at the map $j : A_{n} \rightarrow Z_{n}A = B_{n}A$. I feel that construction of $h$ will be centered around $j$, and that $h$ is exactly the composition $c \circ j$, where $c : B_{n}A \rightarrow A_{n+1}$ is a homomorphism such that $d \circ c = 1_{B_{n}A}$, that is, $d$ is a split epimorphism. But it's hard to construct such $c$, and I'm not even sure if $d$ is indeed a split epi. Am I on the right track? Am I doing this wrong?

I don't think $d$ is necessarily  a split epimorphism: for example, consider the SES of the trivial extension $$0 \to \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \to 0.$$
Any monomorphism into a free abelian group splits, so both assumptions for this chain complex are fulfilled, but $\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ is of course not a split epi.
From your suggestion I guess that you mean the splitting of $d: A_n \to \operatorname{im}d$, which you indeed have: the exact sequence $0 \to Z_{n+1}A  \to A_{n+1} \xrightarrow{d} B_{n}A \to 0$, is due to the acyclicity the same as  $0 \to Z_{n+1}A  \to A_{n+1} \xrightarrow{d} Z_{n}A \to 0.$ Since the left arrow splits, the right one also spilts via some section $s$. Composing $s$ with $A_n \to Z_nA$ gives you a map $A_n \to A_{n+1}.$ Take it from here.
