Let’s consider a rolling cylinder on an inclined plane with some friction. We knew that $\alpha R=a$, where $\alpha$ is the angular acceleration of each constituent particle and $R$ is the radius of disc, so $a$ will be the acceleration of each constituent particle during the rotation and depends on their distance from the center of cylinder.
My question here is how can the acceleration $a=\alpha R$ be equal to the acceleration of center of mass down the inclined plane?
In rotational dynamics, $a=\alpha r$ represents the relationship between angular acceleration and tangential acceleration of a point particle.
In rolling motion, we can analyze it in two motions; translational and rotational. Let $v$ be the translational velocity of com and $\omega$ be the angular velocity. If we consider the point that touches the plane, it has $v$ velocity to the forward and $v'=\omega r$ tangential velocity to the backward. So there net velocity is $v-v'$. If this is a rolling without slipping motion, that particle has zero net velocity. Therefore $$\begin{align}v-v' &=0\\v-\omega r &=0\\v &=\omega r\end{align}$$ If you differentiate this w.r.t. time you will get $a=\alpha r$ whereas $a=\frac{dv}{dt}$ and $\alpha=\frac{d\omega}{dt}$.
You are right -- each point on the cylinder experiences a different acceleration.
When dealing with rotational motion, it's common to break things up into rotation about the center of mass ($\alpha$), and the translation of the center of mass ($a$).
The $a$ you refer to is the acceleration of the center of mass (one point, the center of your cylinder if it's uniform).
Next, $\alpha$ represents the angular acceleration which describes the motion of the edges of the cylinder and how their rotation around the cylinder's center of mass is being accelerated.
Lastly, note that $a=\alpha R$ is a scalar equation, which simply relates the magnitudes of the accelerations.
