Usually, the [𝑟2] factor is justified using one of two methods:

1. The Kinetic Energy Derivation: Substituting (𝑣=𝜔𝑟) into (𝐾𝐸=##\frac{1}{2}𝑚𝑣^2##) yields (##\frac{1}{2}(𝑚𝑟^2)𝜔^2##). While mathematically elegant, this simply groups variables; it doesn't give a dynamic, force-based intuition as to why the resistance scales with the square of the distance.
2. The torque derivation: Using (𝜏=𝑟𝐹) and (𝑎=𝛼𝑟) to get (##\tau=mr^2\alpha##). I got a reasonable explanation for ##𝑟^2## explanation from somewhere but it require prior understanding of the torque. I have still stated the explanation below When you try to rotate an object (like spinning a ball on a string), the distance from the pivot point ([𝑟]) works against you twice:
   • First Factor of [r] (Leverage/Torque): To make something rotate, you apply a force. However, what actually causes rotation is torque ((𝜏=𝐹×𝑟)). If a mass is twice as far away ([2𝑟]), the force we apply has less "rotational leverage" over it. You have to push twice as hard just to apply the same rotational action to that distant mass.
   • Second Factor of [r] (Linear Speed): For a wheel to spin at a certain angular speed (RPM), the outer edge has to travel much faster than the inner part. If a mass is twice as far away ([2r]), it must travel along a circle that is twice as large. To keep up with the rotation, it has to move with twice the linear velocity ((𝑣=𝜔𝑟)). Accelerating a mass to twice the linear speed requires twice the effort. Because these two factors happen at the same time, they multiply together to give ##𝑟^2## dependence

Finally, is there a way to understand 𝑟2 dependence without going circular, or by at least defining one of the things properly without using layman terms like rotational analogues and then throwing up formula. I mean a definition of something should be such that it is self explanatory and directly links to mathematical expression or it lays method by which we can derive the expression

Assume a rigid body rotates about a fixed point that coincides with the origin ##O## of an inertial coordinate frame. The rigid body consists of particles of masses ##m_1,\ldots,m_N## with position vectors ##\boldsymbol r_1,\ldots, \boldsymbol r_N##. The angular momentum of the rigid body about its fixed point ##O## is$$\boldsymbol K_O=\sum_i m_i\boldsymbol r_i\times \boldsymbol v_i.$$Using the formula ## \boldsymbol v_i= \boldsymbol \omega\times \boldsymbol r_i##, we have$$\boldsymbol K_O=J_O \boldsymbol \omega$$where
$$J_O \boldsymbol\omega=\sum_im_i \boldsymbol r_i\times( \boldsymbol \omega\times \boldsymbol r_i)$$
is the inertia operator about the point ##O##.

The moment of inertia of the body about the axis passing through the point ##O## parallel to the unit vector ##\boldsymbol e## is defined as ##I=(\boldsymbol e,J_O\boldsymbol e).##

One just need to understand how the very concept of the moment of inertia arises.

Assume a rigid body rotates about a fixed point that coincides with the origin ##O## of an inertial coordinate frame. The rigid body consists of particles of masses ##m_1,\ldots,m_N## with position vectors ##\boldsymbol r_1,\ldots, \boldsymbol r_N##. The angular momentum of the rigid body about its fixed point ##O## is$$\boldsymbol K_O=\sum_i m_i\boldsymbol r_i\times \boldsymbol v_i.$$Using the formula ## \boldsymbol v_i= \boldsymbol \omega\times \boldsymbol r_i##, we have$$\boldsymbol K_O=J_O \boldsymbol \omega$$where
$$J_O \boldsymbol\omega=\sum_im_i \boldsymbol r_i\times( \boldsymbol \omega\times \boldsymbol r_i)$$
is the inertia operator about the point ##O##.

The moment of inertia of the body about the axis passing through the point ##O## parallel to the unit vector ##\boldsymbol e## is defined as ##I=(\boldsymbol e,J_O\boldsymbol e).##

Since ##\theta## doesn't show in the Lagrangian, it is a cyclic coordinate. So we immediately have the conservation law $$\frac{d}{dt} \frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} m r^2 \dot \theta = 0$$

This is where the ##r^2## factor comes from.

Note that this is the Lagrangian of a free body. There is no force involved. The ##r^2## is not from the force but from the conserved angular momentum itself.

The next step is to use Newton's second law to show how this new quantity changes if a particle is subject to a force.

Try working this out yourself.

In mathematics, moments of a function are certain quantitative measures related to the shape of the function's graph.

For example, in physics, if the function represents mass density,
then the zeroth moment is the total mass,
the first moment is the center of mass (multiplied by total mass), and
the second moment is the moment of inertia.

In statistics, the function is a probability distribution,
then the first moment is the expected value,
the second central moment is the variance,
the third standardized moment is the skewness, and
the fourth standardized moment is the kurtosis.

also useful:
https://en.wikipedia.org/wiki/Moment_of_inertia

I would use a Lagrangian approach for this. For a free body in cylindrical coordinates the Lagrangian is: $$L = \frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 + \dot z^2 \right)$$

Since ##\theta## doesn't show in the Lagrangian, it is a cyclic coordinate. So we immediately have the conservation law $$\frac{d}{dt} \frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} m r^2 \dot \theta = 0$$

This is where the ##r^2## factor comes from.

The conservation of angular momentum is there due to the nature (isotropy) of the Euclidean space, the coordinates are bookkeeping.

For example:
- Charge is what gets affected by an electric field.
- An electric field is what affects charges.
Sounds circular? Who cares? As long it's self-consistent and useful to make predictions it's perfectly fine.

.... force-based intuition ....

...it require prior understanding of the torque. ...

Now consider a particle rotating about a point. Both the speed of the particle and the torque exerted on it scale with the distance ##r## from the axis of rotation.

That is, the further from the axis, the greater the linear speed; with the speed increasing in proportion to ##r##. And as discussed above, the further from the axis, the greater the torque; with the torque increasing in direct proportion to ##r##.

In the meantime, don’t get too bogged down with such questions. It is good to have them but not good to get stuck on them. Circularity is not a real problem. Usually it appears because one part of physics takes ##X## as a definition and derives ##Y## while another takes ##Y## as a definition and derives ##X##. There is nothing wrong with that, and other circularities are also similarly not problematic.

So definitely ask such questions, but don't wait for the answers before continuing your studies.