Conservation of angular momentum in rolling cones 
Consider the situation shown in the figure, A cone is set rolling on a cone so there is angular momentum about the vertical and there is an instantaneous angular momentum that is consistently moving in the horizontal plane.
No forces other than gravity, the normal force, and friction are present (but is set rolling without slipping).

Similarly for a cone undergoing rolling on a plane surface about its apex which most probably has the same explanation.
My question is how gravity and the normal force provide the needed torque to consistently change the angular momentum in the horizontal plane.
Which force dominates and provides the torque also why it only dominates when an initial impulse is given and does not spontaneously start rolling due to the imbalanced forces.
 A: 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.
A: 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.
A: 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).
A: When the motion of a rigid body is known, the equations of dynamics
allow the calculation of the (resulting) forces acting on the body

The contact line of the cones is the axis of instantaneous rotation of the mobile cone and we have the following kinematic relationship:
$$ \overline{\omega} = \overline{\omega}_k + \overline{\Omega} 
\mspace{8cm} (1)$$
$\overline{\omega}  =$ instantaneous angular velocity;
$\overline{\omega}_k=$ spin of the rolling cone;
$\overline{\Omega} =$ precession angular velocity.
On the rolling cone acts the weight applied in its center of gravity G and a system of forces distributed along the contact line.
This system of forces is equivalent to a resultant $\overline{T}$ (a priori unknown) applied in a point A (a priori unknown).
The first equation of rigid body dynamics
$$ {d\overline{Q}\over dt} = \overline{R}  \qquad\to\qquad  -M\Omega^2 \overline{r}_G = M \overline{g} + \overline{T} 
\mspace{2cm} (2) $$
provides the constraint reaction $\overline{T}$.
It is worth to note that $\overline{T}$ lies in the plane of the figure!
The application point A of $\overline{T}$ is given by the second equation (referred to the fixed point O):
$$ {d\overline{L} \over dt} = \overline{M}_o  \qquad\to\qquad \overline{L}= I_1 \overline{\Omega} + I_3 \overline{\omega}_k 
\qquad \overline{M}_o = \overline{r}_G \times M\overline{g} + \overline{r}_A \times \overline{T} \qquad (3)$$
The resulting moment $ \overline{M}_o$ of all forces (with respect to O) is perpendicular to the plane of the figure!
Since in a regular precession $\overline{\Omega}$ is constant,  we have
$$ {d\overline{L} \over dt} = {d(I_3 \overline{\omega}_k )\over dt} = I_3 \overline{\Omega} \times \overline{\omega}_k 
\qquad\to\qquad I_3 \overline{\Omega} \times \overline{\omega}_k = \overline{r}_G \times M\overline{g} + \overline{r}_A \times \overline{T} \qquad (4)$$
It's so possible to calculate the distance $ r_A = \overline{OA}$ (which identifies the point of application A of $\overline{T}$).
From an intuitive insight, it must be understood that the vertical component of $\overline{T} $
balances the weight $M\overline{g}$, while the horizontal component of $\overline{T}$ causes the centripetal acceleration
of the center of gravity G of the mobile cone.
The resulting moment $ \overline{M}_o$ of all forces (referred to O) is perpendicular to the figure and determines the precession motion of the figure axis of the mobile cone.
A: 
The only force that produce a torque to rotate the horizontal cone about the y-axis is the constraint force due to  the rolling condition, not the normal force.
you can obtain the constraint force $~F_c~$ with these equations:
$$I_1\,\ddot{\varphi}_z=F_c\,r_1+\tau_z$$
$$I_2\,\ddot{\varphi}_y=-F_c\,r_2$$
and the rolling condition
$$r_1\,\varphi_z=r_2\,\varphi_y$$
you have three equations for the three unknowns $\ddot{\varphi}_z~,\ddot{\varphi}_y~,F_c$
$$F_c=-{\frac {I_{{2}}r_{{1}}\tau_{{z}}}{{r_{{1}}}^{2}I_{{2}}+{r_{{2}}}^{2}I
_{{1}}}}
$$
and the torque about the y-axis if $~F_c\,r_2$
A: Intuitively, to me this is a gyroscope that is constrained from falling or slipping.  A simple gyroscope  that is not spinning rapidly, constrained about only one point, will fall as it tries to precess.  Here the initial rotation of the top cone produces an initial angular momentum; the torque from gravity causes a change in angular momentum resulting in precession while the constraint from the lower cone prevents falling downwards or slipping off the lower cone.  Not sure about the torque from friction.
