I was wondering how the stationary flywheel scenario worked.
As Adrian said if the flywheel were not to spin, it would fall over (i.e. you would not observe what is called precessional motion). Why? Because there's no angular momentum due to the spin of the flywheel and thus no torque.
Why is it that the gyroscopic falls along a "circular path?"
This is basically because of the spinning of the flywheel. Let's say that it is spinning clockwise. Then by convention $\vec L$ points outwards along the symmetry axis.
The key is this: the weight of the flywheel exerts a torque on the gyroscope about the pivot. Such a torque changes the direction of the angular momentum, triggering precessional motion.
By the right hand rule you can guess that the direction of the torque is perpendicular to the plane in which the weight and O-CM distance vectors lie
how is angular momentum conserved in this case, considering Fg is a conservative force?
The angular momentum of the gyroscope system is not conserved. This is due to the presence of an external torque:
$$\vec \tau = (m \vec g)\vec r$$
Where $m$ is the mass of the flywheel and $\vec r$ is the O-CM distance.
Let me give you an example in which the angular momentum of the system is conserved.
Imagine a rocket containing a gyroscope that is initially not rotating (and it is in the space). Then, the total angular momentum of the system is zero at this point. Say you want the rocket to spin left. Then, if you make the gyroscope rotate clockwise (by remote control) you will generate an angular momentum pointing outwards. As the gyroscope-rocket system is isolated, angular momentum must be conserved. Thus, the rocket must rotate counterclockwise and generate an angular momentum pointing inwards so that both angular momentum vectors cancel each other out.