This alone does not explain why they did so. But this is all assuming that the trajectory is planar. If there was a plane change coupled with the second burn at high apogee, it could explain it.
If there is a plane change involved (as apparently there is), then getting further out before executing the plane change involves less delta-v; performing a plane-change maneuver at apogee is by far the most efficient way.
Comments
Bi-elliptic transfer sounds like a plausible explanation but my math gives it a bigger delta-v requirement than a straightforward Hohmann transfer.
Bi-elliptic transfer: (2.713240 + 0.908982 + 0.485255) = 4.107477 km/s Hohmann transfer: (2.335977 + 1.431307) = 3.7672853 km/s
This alone does not explain why they did so. But this is all assuming that the trajectory is planar. If there was a plane change coupled with the second burn at high apogee, it could explain it.
If there is a plane change involved (as apparently there is), then getting further out before executing the plane change involves less delta-v; performing a plane-change maneuver at apogee is by far the most efficient way.