Origins of Moment of Inertia

The equation you are referring to is the expression for the moment of inertia of a point particle of mass $m$ at a distance $R$ away from some axis. This expression is really the definition of the moment of inertia for a point mass, so the question becomes "where does this definition come from, and why is it useful?"

Well, for simplicity's sake, suppose that such a point mass is rotating around the aforementioned axis on a string, then if a tangential force $F_t$ is applied to it, its resulting tangential acceleration $a_t$ will satisfy $$ F_t = ma_t $$ so the torque on it is $$ \tau = Rma_t $$ On the other hand, recall that the tangential acceleration $a_t$ and angular acceleration $\alpha$ are related by $a_t = R\alpha$, and plugging this into the right hand side of the expression for the torque gives $$ \tau = mR^2 \alpha $$ Notice that the quantity $mR^2$ has magically appeared. Well clearly it's not actually magic; the point really is that if we think of torque as a sort of rotational analog of force, and angular acceleration as the rotational analog of acceleration, then this shows that the quantity $mR^2$ is a sort of rotational analog of mass for a point mass. As a result, we give it a special name: moment of inertia.

It's important to point out that although I used the example of a point mass undergoing uniform circular motion to motivate the definition of moment of inertia, there are significantly more involved and general derivations that lead to a quantity called the inertia tensor which is the generalization of the moment of inertia for non-pointlike bodies undergoing arbitrary rotation.

See, for example, the following answer:

https://physics.stackexchange.com/a/89304/19976

It looks arbitrary because the concept of Inertia is defined that way. If someone can digest $F=ma$ (linear), the same should be followed for $\tau=I\alpha$ (angular)

Moment of Inertia is just the rotational substituent for the inertia coefficient (mass) in linear motion. It's not some kind of arbitrary value. It's actually given by $I=Kmr^2$. The $mr^2$ normally seen everywhere is approximated only for a point mass because we always consider point particles to derive something. In case of rigid bodies, we go throught the derivation stuff, by choosing certain axis of rotation, make it spin and finally plug it into our formula $I=r^2\int dm$

So, it can come in various delicious flavors like - for $K=\frac{1}{3}$, we get the moment of inertia for a rod where $r$ acts as the distance from the axis of rotation (which is at one end) with respect to the mass $m$ of the rod. Different derivations "do" exist for different rigid body systems...