What is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?

As mentioned above, there is no standard notion of "orientability" for CW-complexes. I will assume you meant manifolds instead.

• The open Möbius band is a nonorientable manifold. It has the homotopy type of a circle, hence its homology groups are free abelian.

• Let $M$ be a compact, non-orientable $n$-manifold. By Poincaré Duality, $\mathbb{Z}/2\mathbb{Z}$-Poincaré Duality and the Universal Coefficient Theorem, both $H_1(M,\mathbb{Z})$ and $H_{n-1}(M,\mathbb{Z})$ have nontrivial $2$-torsion.

• $\mathbb{R} \mathbb{P}^3$ is orientable but has first homology group $\mathbb{Z}/2\mathbb{Z}$.
• A nonorientable manifold admits a connected $2$-sheeted orientation covering. In particular its fundamental group admits an index $2$ subgroup, so for instance any simply connected manifold must be orientable. This does not mean that there must be an order $2$ element in the fundamental group of a nonorientable manifold: e.g. it follows from basic group cohomology that the fundamental group of any manifold with contractible universal cover -- so in particular, every nonorientable closed surface other than the projective plane -- has torsionfree fundamental group.

Regarding the assertion in wikipedia, recall that a compact connected manifold (without boundary) of dimension $n$ is orientable if and only if the top homology group $H_n(X;\mathbb Z)\cong \mathbb Z$ (see in Hatcher's Algebraic topology the section entitled Orientations and homology, chapter 3.3).