This is no different in GR than in SR. Vectors like the momentum four-vector live in the tangent space at a particular spacetime point, which means they act just like you expect from SR.
I don't think it equals the same thing as the special relativistic case, because the time component isn't related directly to the energy anymore (is this correct??).
No, it's the same as in SR. You can define a local frame of reference, which defines a unit vector in the time direction.
The gamma factor ultimately comes from flat space metric and so is not appropriate here.
No, the gamma doesn't have anything to do with the metric. If you choose appropriate local Minkowski coordinates, then the metric looks exactly like the metric in SR, diag(1,-1,-1,-1). The gamma just occurs because you're normalizing the velocity vector.
Since you mention a geodesic, it may be that you're thinking of the case where the spacetime has a timelike Killing vector, and then there is a conserved energy. That's a different thing. Most spacetimes don't have a timelike Killing vector, and then there is no conserved energy that you can define for test particles. But that has nothing to do with the general ideas of how you define the energy and the momentum four-vector, which are the same in all cases.
It's also not true that the momentum four-vector is defined from a normalized velocity vector. That wouldn't work for massless particles (in either SR or GR). The only real things you can say in general are that the mass is the norm of the momentum vector, and the momentum vector has to be parallel to the (normalized or unnormalized) velocity vector (or else it would violate rotational symmetry).