Warning--stream of consciousness follows...
I don't have a good enough handle on gyroscopic concepts to be able to intuitively think through the spin-stabilized case, but I can think through the aerodynamically stabilized case. I'm going to start from there and then see if that gives me any insight into the spin-stabilized case.
In the aerodynamically stabilized case, it's largely a matter of where the center of mass and the center of pressure (due to drag) are located on/in the object. If they are co-located (or close to being co-located) on/in the object, the object will tend to tumble, if they are significantly separated, the object will tend to want to fly with the center of mass forward and the center of pressure rearward.
If that concept is difficult to visualize, imagine dropping a straw that has a paperclip attached to one end. The paperclip end will consistently hit the ground first no matter how you orient the straw before you let go of it. The center of mass is near the paperclip end but the center of pressure (focus of drag) is somewhere in the middle of the length of the straw. So the drag exerts force on the straw and that means the center of mass ends up on the "front" end of our projectile.
Now let's change things up by increasing the length significantly without significantly affecting the center of mass. That would change the center of pressure and could potentially affect stability depending on where the length is added. If it's added toward the rear, it would be expected to improve stability, if it's added toward the front, it would be expected to make the object less stable.
Think of taking a longer straw and attaching the paperclip so that the same amount of straw trails it as in the first example but now there's some straw extending forward of the paper clip. Assuming that there's enough added forward length, the straw will now flutter or tumble when it's dropped. The center of pressure has been moved forward somewhat relative to its initial position with respect to the center of mass. That's because the paperclip is still significantly heaver than the straw and therefore the center of mass will still be very close to where the paperclip is. But now the straw extends forward of the paper clip, so even though the center of pressure (focus of drag) is still near the middle of the length of the straw, that center of pressure is now farther forward with respect to our center of mass.
So, let's try to extend that to the spin-stabilized case. We undertand that the spin-stabilization is there to compensate for the fact that the projectile isn't inherently stable (either the center of mass and center of pressure are too close together or the center of mass is actually rearward of the center of pressure).
Earlier, with our straw experiment, we established the idea that extending the length of the projectile forward without changing the mass of the projectile significantly would have a tendency to move the center of pressure forward while leaving the center of mass nearly unchanged. The effect of the change was that it made the projectile less stable. We would expect a less stable projectile to be harder to stabilize and therefore we would logically expect that we might need more spin to achieve the stability we want.
Getting back to our specific question, extending the length of the bullet in the forward direction without significantly changing the weight seems like it would make the projectile less stable and that suggests that we would want our formula to reflect that change so we can calculate an accurate twist to stabilize it.