I know of the following proposition:
Let $p: (N, h) \rightarrow (M,g)$ be a Riemannian covering map. The geodesics of $(M,g)$ are the projections of the geodesics of $(N,h)$, and the geodesics of $(N,h)$ are the liftings of those of $(M,g)$.
If I would be asked to prove this, I would use the local length minimizing property of geodesics and the fact that the covering map doesn't change the length of a smooth curve since it is a local isometry. However, in the book "Riemannian Geometry" by Gallot, Hulin and Lafontaine, this fact is proven way before the local length minimizing property is discussed (Proposition 2.81). See also this question for a proof: geodesics, covering map and its lift
I'm not sure I fully understand (the final step of) these proofs. Would the following argument be correct?:
If one considers local coordinates, a geodesic is a smooth curve that fulfills a certain differential equation which essentially only depends on the Christoffel symbols, which in turn only depend on the Riemannian metric. Since the Riemannian metric behaves nicely under a Riemannian covering map (i.e. $h = p^*(g)$) the above statement follows.
EDIT: Just to clarify, I understand how the proposition can be proven using the locally length minimizing property of geodesics. I want to understand how it can be proven without using this property of geodesics.
sFrom the length point of view, one needs to remember that there is no much difference between the set of paths in $N$ and in $M$. More precisely, if we fix a base point $x_0$ in $M$ and $y_0$ in $N$ s.t. $p(y_0)=x_0$ the natural projection $p_*$ induces a bijection between the set of path issuying from $y_0$ to the set of paths issuying from $x_0$, and this bijection preserves lengths. In particular critical points of the lengths functions, i.e. geodesics issuying from $x_0$ and $y_0$ are "the same", i.e a projection of a geodesics upstairs (resp. a lift of a geodesic downstairs) is a geodesic.