The answer to Question 1 is no.
Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$.
Since $B$ is cyclic of order $4$, if it were contained in a proper direct summand of $A$ then it would be contained in a cyclic subgroup of $A$ of order $8$, and so $(2,1)$ would be equal to $2a$ for some $a\in A$, which it's not.
In fact, if you consider only the case where $A$ has exponent dividing $p^n$ for some prime $p$ and natural number $n$, there are infinitely many indecomposable such couples if $n=6$, and the classification is in some sense "wild" if $n\geq7$. See, for example,
Ringel, Claus Michael; Schmidmeier, Markus, Submodule categories of wild representation type., J. Pure Appl. Algebra 205, No. 2, 412-422 (2006). ZBL1147.16019.
and its references.