Centripetal force when switching between frames 
Cycloid is a curve which can be defined as a trajectory of a point marked on the rim of a rolling wheel or radius
R. Determine the curvature radius of such curve at its highest
point. from page-11 of Jaan Kalda's notes
Solution:

As the wheel is rolling, we have that $\omega=\frac{v}{R}$.
The speed of the highest point in the lab frame is
$$v+\omega R=2v$$
Therefore, we find that the centripetal force at the highest point is
$$a_c=\frac{(2v)^2}{r}$$
The speed of highest point in frame of wheel's centre is $\omega R=v$.
Therefore, the centripetal force in the wheels's center is
$$a_c=\omega^2R=\frac{v^2}{R}$$
As both frames are inertial frames,
$$\frac{v^2}{R}=\frac{4v^2}{r}\Longrightarrow\boxed{r=4R}$$

Page-25 of this pdf
In this post, I found that centripetal force depends on the coordinate system where you write the equations in, but it in the solution, it was used that centripetal force is same in all frames. So, what made it that we could equate centripetal force of two different case in this case?
 A: 
In this post, I found in centripetal force depends on the coordinate system where you write the equations in, but it in the solution, it was used that centripetal force is same in all frames. So, what made it that we could equate centripetal force of two different case in this case?

In the cited post it says "both frames do agree on the magnitude and the direction of the acceleration itself. So it is not that the force disappears, just that calling the force 'centripetal' no longer makes sense".
So you can always equate the force in the two frames. Forces are always the same in all inertial frames.  It is just the label "centripetal" which does not apply in both frames. Again, that is because the motion is not uniform circular motion in both frames. So the force is the same, just not the label "centripetal".
Note that this does not contradict Jaan Kalda's notes. If you read his example carefully you will see that he is careful to not call the force "centripetal" in the frame where the path is a cycloid. Instead, in that frame he is only calculating the curvature. In his words:

The formula $v^2/r$ can be also used for the perpendicular to the motion acceleration of a point along a curved trajectory; then, $r$ is the curvature radius of the trajectory.

So the formula can be used for different uses in different frames. In a frame where the motion is uniform circular motion that gives the centripetal acceleration. But in any frame it can be used to determine a radius of curvature. In the frame where the motion is uniform circular motion this radius of curvature is constant and equal to the radius of the circle. In the cycloid frame the radius of curvature is not constant and the calculation is only for that specific instant.
