If you define kinetic energy as $KE =\frac{1}{2} m (\vec{v}_G \cdot \vec{v}_G) + \frac{1}{2} I_G (\vec{\omega} \cdot \vec{\omega} )$ where $v_G$ is the linear velocity of the CG and $I_G$ is the mass moment of inertia about the CG, and then transform the quantities to the handle of the rigid body (i.e. the pin) then you will get what Wikipedia has.
Moments of inertia are additive. Suppose you have particle $A$ with a moment of inertia $I_A$ and particle $B$ with a moment of inertia $I_B$. Then the total moment of inertia of both particles is just $I_A + I_B$.
You can imagine a ring as being made up from lots of point particles, all at a distance $R$ from the central axis. In that case the moment of inertia of the ring would be the sum of all the moments of inertia of all the point particles:
$$ I_{ring} = I_A + I_B + \, ... = m_A R^2 + m_B R^2 + \, ... $$
Since $R$ is a constant we can take $R^2$ outside the sum to get:
$$ I_{ring} = \left(m_A + m_B + \, ...\right)R^2 $$
and $m_A + m_B + \, ...$ just adds up to the total mass of all the particles $M$. so we end up with:
$$ I_{ring} = MR^2 $$
This is a special case of a general principle in calculating moments of inertia. For any body of any shape we can divide it up in to infinitesimal elements then add up their moments of inertia to get the total. In general this is done by integration, as Wet Savanna Animal explains in his answer.
Best Answer
A hollow sphere will have a much larger moment of inertia than a uniform sphere of the same size and the same mass.
If this seems counterintuitive, you probably carry a mental image of creating the hollow sphere by removing internal mass from the uniform sphere. This is an incorrect image, as such a process would create a hollow sphere of much lighter mass than the uniform sphere. The correct mental model corresponds to moving internal mass to the surface of the sphere.