FALSE: every finitely generated module over a domain is a direct sum of cyclic modules

Let $R=\mathbb{Z}[x]$ and $I=(2,x)$. The ring is a domain, and $I$ is generated by the displayed elements, so it is a two-generated torsion-free $R$-module.

Any nonzero cyclic submodule of the torsion-free ideal is isomorphic to $R$. If a direct-sum decomposition contained two nonzero cyclic summands with nonzero generators $a,b\in I$, the relation $b\cdot a-a\cdot b=0$ would be a nontrivial relation between them, contradicting directness. Thus a cyclic decomposition could have at most one nonzero summand.

A single nonzero cyclic summand would make $I$ principal by [F1]. But a generator would divide both $2$ and $x$ in $\mathbb{Z}[x]$, hence would be a unit; that would give $I=R$, while reduction modulo $(2,x)$ shows $1\mathrm{\notin }I$. Thus $I$ is not cyclic and has no direct-sum decomposition into cyclic modules, refuting the claim and isolating the PID hypothesis.