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Typo! Sorry, meant eigenvector. One eigenvector is $(1,0)$ corresponding to infinity. The ratio of the entries of the other eigenvector gives the other end of the geodesic. Since the matrices commute they have common eigenvector and hence common fixed points on $CP^1$ which are the endpoints of the geodesic.
No, this won’t hold in general. One may construct manifolds with $\pi(N)$ of unbounded diameter and $\pi(\partial N)$ bounded diameter. I might try to write a more complete answer, but to summarize: one may take a hyperbolic handlebody with convex core $N$ and $\partial N$ of bounded diameter but $N$ of arbitrarily large diameter. Perturb a bit so that one may extend by a reflection group using Thurston’s reflection trick, then a manifold cover will have $N$ embedded and hence won’t satisfy your conditions.
I think you tube/add 1-handles as much as possible. If this doesn’t give a connected preimage, then the manifold maps to a proper subgroup of Z, a contradiction.
