It's not about what causes the gravitational force to be greater. It's that the normal force will be less (these two statements are the same mathematically, but colloquially they change the focus).
Let's look at a simpler example. Take a box and put it on the floor. We can say the normal force acting on the box is equal to its weight. Now let's apply an upward force to the box (for example, with your hands) with a magnitude less than the box's weight. What happens? The normal force will decrease to be less than the box's weight even though the box will remain at rest.
Now let's go to your example. Because the block is undergoing circular motion, we know that the normal force has to be less than the gravitational force to have a net inward force. The easiest way to relate this to the previous example is to move to a frame that is rotating with the block. In this frame, the centrifugal force acts like the force from our hands on our box. The block will stay at rest (in the rotating frame), but the normal force will be less than the weight.
The main source of confusion here is from your more general thought
I have always known the normal force to equal the resultant force acting against it, which in this case is gravity.
It is time to abandon this. It is true in most introductory physics problems when you are first introduced to the normal force, but this is not true in general. The normal force is only equal to the "forces against it" if there is no acceleration in the direction of the normal force, since then we can say that $N=F_\text{oppose}$. If there is acceleration along the direction of the normal force, then there is no longer a balance of forces in that direction. Your example here shows this. A simpler example would be the case of accelerating in an elevator, where the normal force is larger (smaller) than the weight if the elevator is accelerating upwards (downwards). A more complicated example that you might have seen where the normal force is not equal to the forces "against it" is an object going around a banked curve, where the normal force is perpendicular to the curve, but the acceleration has a non-zero component perpendicular to the curve.
