As noted, and as you know, every subgroup of an abelian group is normal; so the socle of an abelian group $G$ is generated by the minimal nontrivial subgroups.
If $H$ is a minimal nontrivial subgroup of $G$, and $h\in H$, $h\neq 1$, then $\langle h\rangle = H$; thus, $H$ is cyclic, and has no nontrivial subgroups. Therefore, $H$ is cyclic of prime order. Conversely, if $H$ is a subgroup of prime order, then it is necessarily minimal nontrivial, since it has no proper nontrivial subgroups.
So the socle must be generated by the elements of prime order. In particular, $(\mathbb{R},+)$, being torsionfree, has trivial socle, as does every torsionfree abelian group.
Thus, if $G$ is abelian, then $\mathrm{socle}(G)=\mathrm{socle}(G_{\rm tor})$, the torsion subgroup of $G$. And
$$\mathrm{socle}(G) = \langle x\mid x\text{ has prime order}\rangle.$$
You can get a bit more information (depending on how much Robinson has proven). Since the torsion subgroup is the direct sum of its $p$-parts, you can just look at the separate $p$-parts, and then just look at what some authors denote $G[p]$, the set of elements of order dividing $p$. So
$$\mathrm{socle}(G) = \bigoplus_{p\text{ prime}}G[p].$$
(Your final example comes from the fact that $\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_3$, so you "really" have $\mathbb{Z}_2\times\mathbb{Z}_3^2$...)
As Ycor points out this question is easy: A torsion free group (Wang or otherwise) is abelian if and only if its commutator subgroup is finite.
For the first direction, abelian implies trivial commutator subgroup so certainly the commutator subgroup is finite.
Reverse impliciation: since $ G $ is torsion free then any finite subgroup is trivial. So if the commutator subgroup is finite then actually the commutator subgroup is trivial and $ G $ is abelian.
Best Answer
Yes, this is the same thing as an $\mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $\mathbb{F}_p$ is the quotient ring $\mathbb{Z}/(p)$, an $\mathbb{F}_p$-module is the same thing as a $\mathbb{Z}$-module in which every element is annihilated by $p$. But a $\mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.