A rigid body revolves around some origin with angular speed $\Omega$. Then its kinetic energy is (parallel-axis theorem)
$$K = \frac{1}{2}I\Omega^2 = \frac{1}{2}md^2\Omega^2 + \frac{1}{2}I_\text{cm}\Omega^2$$
My question is, conceptually, why should this second term there if the object isn't necessarily also rotating about its center of mass with angular speed $\Omega$? Why should a rigid body have more kinetic energy for the same $\Omega$ than a point particle of identical mass?
Because part of the objects mass is further away or closer to the center of rotation. Since the contribution to the kinetic energy is quadratic, not linear, in the distance from the axis of rotation, you don't get the same answer for a point particle moving along the center of mass of the object. That discrepancy is accounted for by the second term.
The kinetic energy (KE) of a system of particles, of which a rigid body is a special case, is the kinetic energy of the center of mass (CM) plus the KE of the system with respect to its center of mass. This is proven in many physics mechanics textbooks, such as Classical Mechanics by Goldstein.
