Let $F$ be a free abelian group (of finite rank). For every two direct summands $A$ and $B$ of $F$, either $A$ is a direct summand of $B$ or $B$ is a direct summand of $A$ up to isomorphism.
My question is:
Which abelian groups have this property other than free abelian groups?
If $G$ is a finitely generated abelian group then the classification of finitely generated abelian groups can be used to classify all such groups.
Write
$$G\simeq\mathbb{Z}^n\oplus F$$
where $F$ is finite. If $F$ is non-trivial (i.e. $G$ is not free) and $n>0$ then obviously this group cannot have this property since $\mathbb{Z}^n$ cannot be a direct summand of $F$ and vice versa.
So we are left with the case when $G$ is finite. If $|G|$ is divisible by two different primes then $G\simeq H\oplus K$ with $\gcd(|H|,|K|)=1$ and hence neither $H$ can be a summand of $K$ nor vice versa.
Thus we are left with $G$ being a $p$-group. You can write $G$ as
$$G\simeq\mathbb{Z}_{p^{a_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{a_m}}$$
Now every component is indecomposable and thus it can be easily seen that (by the uniquness of decomposition) that the only way for $G$ to have the desired property is when $a_1=\cdots=a_m$. And indeed every such group has the desired property.
I cannot help you in general abelian case.