$\newcommand{\Z}{\mathbb Z}$
Question: Let $m$ and $n$ be positive integers. What are all the abelian groups $A$ such that there is a short exact sequence $0\to \mathbb Z/p^m\mathbb Z\to A\to \mathbb Z/p^n\mathbb Z\to 0$.
It is clear that any such abelian group $A$ has cardinality $p^{m+n}$. Thus by Fundamental Theore of Finitely Generated Abelian Groups we have $A\cong \Z/p^{k_1}\Z\oplus \cdots\oplus \Z/p^{k_r}\Z$ for some $k_1\leq \cdots \leq k_r$. Also, since $A$ contains a copy of $\Z/p^m\Z$, a cyclic group of order $p^m$, we must have $k_r\geq m$. I am unable to see what to do next.
Also, I am particularly interested in the solution given here in the form of Proposition 0.1 on pg 2. On the first line of the solution it uses a notion of "pull-back" which I do not understand. Can somebody shed some light on this solution?
The solution you reference is unnecessarily sophisticated...but at least it tells you the answer.
You have already found two properties that any such group $A$ must have:
1) $A$ has order $p^{m+n}$. In particular, it is a commutative $p$-group.
2) $A$ has an element of order $p^m$.
You need one more property:
3) $A$ can be generated by two elements.
Thus in your appeal to the structure theorem for finitely generated abelian groups you know you can take $r \leq 2$. The two extreme cases are a cyclic group $\mathbb{Z}/p^{m+n}\mathbb{Z}$ and the group $\mathbb{Z}/p^m \mathbb{Z} \oplus \mathbb{Z}/p^n \mathbb{Z}$. You should be able to check that these groups and "everything in between" can actually occur.