Calculating force through ICOR (instantaneous centre of rotation) 
A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). 
If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ represents angular velocity. 
If I calculate the force using the radius of curvature of cycloid, I end up with the value of $F=mR\omega^2$, which is indeed true. 
My concern is that why doesn't ICOR provide me with the correct answer. 
 A: $ P$ is a non-inertial frame of reference which is why it does not provides us with correct answer. In order to obtain the correct result we need to write the relation between accelerations as $a$$real$ = $a$$0$ +$a'$.  Where $a$$real$ denotes acceleration of the particle(whose motion is considered) in inertial frame, $a$$0$ denotes acceleration of non-inertial frame and $a'$ denotes acceleration of particle in non-inertial frame. 
$a$$0$ is equal to $Rω²$ directed towards centre of the wheel  whereas $a'$ is the acceleration of  the top most point equal to $2Rω²$ directed radially towards the center of the wheel.
*Note: Since both points, ICOR and top most particle are diametrically opposite hence addition of acceleration vectors pointing towards centre produce a cancelation effect.
Adding both vectors results in real acceleration equal to $Rω²$ pointing towards the centre of the wheel, which is the correct result. 

A: The "true" acceleration of the rim point A is $a_A = 2 R \omega^2$. How did you decide the true value is $R \omega^2$?
You can look at the ICOR and arrive at $2 R \omega^2$, or you can add the acceleration of A due to the rotation $R \omega^2$ plus the acceleration of the center of mass $R \omega^2$ to arrive at the same value. 
BTW Taking the acceleration of A and multiplying by the entire mass $m$ of the body is meaningless as a force quantity. Only the acceleration of the center of mass is important in deriving inertial forces. The is because linear momentum is defined as the mass of a body $m$ times the velocity of the center of mass $v_C = \omega R$ and force is the time derivative of momentum.
