Conservation of angular momentum in rolling cones
The normal force acting on the rolling cone along the line of contact must must produce a torque about the apex which is slightly greater than that produced by gravity. This excess torque vector is horizontal and perpendicular to, $v_2$, and causes the associated angular momentum to swing in that direction.
You can separate the movement into the spinning around the cone's axis of symmetry and orbital movement around a vertical axis going through the top of the cone. The orbital angular momentum is constant (as neither the vertical axis, orbital speed or moment of inertia change), but the angular momentum associated with the spin changes direction, as the cone orbits. By drawing everything out it can be checked that the infinitesimal change of this angular momentum is in the horizontal plane and it has the direction opposite to the orbital velocity. This must be the direction of the moment of the forces that cause this change. That means that the forces that cause this must be vertical.
One can also see the cone as a collection of slices, wheels rolling on the surface. Each slice is tilted towards the vertical axis going through the top of the cone. One can analyze the rolling of such tilted wheel and it can be found that it will also change the direction of its movement just like the whole cone does. That is because its center of gravity is not directly above the point where the wheel touches the surface, which cause an imbalance of the momenta of these forces. The fact that a cone is a solid means just that the movement of all slices must be coordinated.
Therefore it can be concluded that it's the gravity and the normal forces. While the forces negate each other, their momenta don't have to. The situation seems to be somewhat analogous to the gyroscope precession except that for the gyroscope all of the normal force is applied to the top, and for the cone it can be distributed along the line of contact. The condition that all the rolling happens without sliding seem to cause an imbalance between the momenta of the the gravitational force and the normal forces that is necessary to change the axis of the spinning as the cone orbits the vertical axis.
We can say from the symmetry of the system, that the vertical component of the angular momentum is constant. So $$\tau = \left(\frac{dL_x}{dt}, \frac{dL_y}{dt}, 0 \right)$$
When the cones are at rest, the system is in equilibrium and there is no torque. So the sum of the reaction forces: (friction + normal) must pass through the center of gravity of the movable cone.
When the movable cone is rotating, the friction force that was necessary to avoid slippage must increase. It acts now also as a centripetal force. That $\delta F$ generates now a binary with the weight at CG.
Checking its direction, we can see that the torque is in the plane $xy$ as required by symmetry.
(It is the same answer of R.W.Bird, I haven't note that part of it was in his comments).