A compactness argument which Hodges may have had in mind can go as follows. Since a subgroup of a totally ordered group is also a totally ordered group, it suffices to embed the given abelian torsion-free group $G$ into a totally ordered group, i.e., to show that the theory of totally ordered abelian groups is consistent with the diagram of $G$. By the compactness theorem, it is enough to show that this is true for any finite subset of the diagram. This finite subset only mentions finitely many constants from $G$, hence it suffices to show that every finitely generated subgroup of $G$ is totally orderable. However, every finitely generated abelian torsion-free group is isomorphic to $\mathbb Z^n$ for some $n\in\omega$, which can be given e.g. the lexicographic order.

A compactness argument which Hodges may have had in mind can go as follows. Since a subgroup of a totally ordered group is also a totally ordered group, it suffices to embed the given abelian torsion-free group $G$ into a totally ordered group, i.e., to show that the theory of totally ordered abelian groups is consistent with the diagram of $G$. By the compactness theorem, it is enough to show that this is true for any finite subset of the diagram. This finite subset only mentions finitely many constants from $G$, hence it suffices to show that every finitely generated subgroup of $G$ is totally orderable. However, every finitely generated abelian torsion-free group is isomorphic to $\mathbb Z^n$ for some $n\in\omega$, which can be given e.g. the lexicographic order.

Notice that only the last step used something specific about abelian groups. The same argument shows that a (nonabelian) group is totally orderable if and only if all its finitely generated subgroups are, and likewise for other ordered structures (e.g., semigroups or rings).