Yes, the kinetic energy of a uniform ring on the outside of the wheel is exactly double of the kinetic energy of an equal mass elsewhere on the vehicle. You can prove this by integrating around the circle; the key is that the ratio of the velocity of a point on the ring to the velocity of the vehicle depends solely on the angle. (The point in contact with the ground isn't moving, while the point at the top of the tire is moving at 2v.)
Note that this applies only to the outside of the wheel; there is a cost to weight closer to the centre of the wheel, but not as much. (Also, things get weird with non-rotationally-uniform wheels, but people tend to avoid those for other reasons.)
https://en.m.wikipedia.org/wiki/Moment_of_inertia
It's discussed in terms of inertia and momentum, but these quantities are the useful quantities when discussing rotation about an axis.
Speaking loosly, applying a force perpendicular to the radius of a rotating body changes its angular velocity, which is a factor of momentum. The moment of inertia determines in part how large the torque on the mass is for a given force about the center.
For racing/road bikes, maybe. You're leaving out the 26" wheel adult bikes (which over time has encompassed 5 different rim diameters[1]) and 24" wheel bikes that are targeted at youth, but smaller adults sometimes prefer (well, those that have been give the option).
My bike pedaled contraption is a rear suspension compact long wheelbase recumbent with two real 20" (406mm rim) tires and they're spectacular.
Is this always true regardless of wheel radius? At first I was thinking it has to depend on radius, but then thought about how the ground contact point has a zero velocity, and the exact other side of the wheel is always double the velocity of the bike frame relative to the ground, regardless of wheel radius. I'm not sure if this is a reasonable intuition. Maybe it makes sense that a gram added to the rim is always double the difference in effort that a gram added to the axle or frame or rider is.