Center of gravity of a ring and static equilibrium

There is not one individual force acting at the center of mass. In a simple, first-order model, multiple gravitational forces are acting downward on every bit of mass in the ring. If we add all those vectors together, the net vector is the mass of the ring, $m$, times the gravitational field strength, $g$. The net line of action passes through the geometric center of the ring (uniform mass distribution).

Each bit of mass in the ring interacts with its nearest neighbors through electromagnetic forces (molecular bonding) so that the ring maintains is shape. The fact that bits of the ring are not accelerating relative to each other tells use the net internal forces are zero, so we don't worry about them. In reality, for a vertical ring, the ring distorts slightly and creates local stresses, but we ignore those in introductory physics.

If the ring is sitting on a table, the table exerts an upward force on the ring at the point of contact. If we observed the ring to not be accelerating (constant velocity or constant zero velocity), we know that (in our chosen reference frame) the vector sum of forces is zero. How can that happen? Because the intermolecular forces of the table hold it together and prevent the ring from passing through it, and result in an upward force on the ring.

The lines of actions of the forces are only important in 1) determining how to add the vectors because of the angular relationships and 2) in determining whether there is any net torque because the lines are or are not co-linear. In your case, you have specified that the normal (electromagnetic) force from the table is co-linear with and opposite in direction to the net gravitational force from the Earth.