I find that an easy, never-fail way to analyze the forces acting on an object in circular motion is:
(forces toward the center) - (forces away from the center) = mv²/r.
mv²/r is the amount of center-directed force necessary to make the object, which constantly wants to move in a straight line, move in a circle instead. It's often called the centripetal force, but that's just a label we put on the combination of identifiable forces like gravity, normal force, tension, etc., that gives us mv²/r.
Here is a force diagram (a.k.a. free body diagram) for any car-on-a-(circular)-hill scenario:
Notice that there are no forces labeled as centripetial force, or mv²/r. Our equation becomes
$$mg(sin\theta) - F_N = mv²/r,$$
where $\theta$ is the angle that the position vector of the car makes with the horizontal, and FN is the normal force exerted on the car by the hill, or road.
At the top of the hill, where $\theta$ is 90°, the equation reduces to
$$mg - F_N = mv²/r$$
So yes, in the case of a car going over a circular hill, we see that the centripetal force is the difference between the car's own gravitational force and the normal force exerted on it by the hill, as you suggested.
If we rearrange the equation: $$F_N = mg(sin\theta) -mv²/r$$
We see that only when v = 0 is the normal force exerted by the road on the car equal to the gravitational force on the car. In that situation they are equal and they oppose each other, so they look like a Newton's Third Law pair. But they are not. Note that they act on the same object.
There are two Newton's Third Law pairs in the situation. One pair consists of the force mg(sin$\theta$), exerted on the car by the mass of the Earth, and an equal force exerted on the mass of the Earth by the car. The other pair consists of FN exerted on the car by the road, and an equal normal force exerted on the road by the car. That pair gets smaller as the velocity increases, as shown by the latest equation above. When FN gets to zero, the car leaves the road tangentially.