This is right: a map $f:A \rightarrow B$ is surjectiv $\Longleftrightarrow$ $f$ has a right-inverse. The proof needs the axiom of choice, as you pointed out correctly. But this is just a map of sets.
EDIT: Here In the following I'm talking about groups (vector spaces, vector bundles, presheaves, sheaves,... should also do the job)
Condition (2) is always satisfied since every
Every short exact seqeunce can be seen as a sequence $$0 \overset{inc_0}\rightarrow A \overset{inc}{\rightarrow} B \overset{\pi}{\rightarrow} B/A \rightarrow 0$$ where $inc$ denotes the inclusion and $\pi$ the projection.
But (as for example Charles Siegel pointed out) surjectivity gives you just a rightinverse $u:C \rightarrow B$ as map of sets. So if you have further structures (let's say a group structure, vector space structure, etc.), this doesn't mean, that the map $u$ is an inverse with respect to these structures
