What is the relation between angular and linear acceleration?

A point, whose position vector is $\vec r$, of a rigid body with angular velocity $\vec \omega$ has velocity $\vec v=\vec\omega\times\vec r$. By differentiating $\vec v$ with respect to time we obtain the acceleration $$\vec a=\vec\alpha\times\vec r+\vec\omega\times\vec v,$$ where $\vec\alpha=d\vec\omega/dt$ is the angular acceleration.

The first term, $\vec\alpha\times\vec r$, is parallel to the velocity vector and is normally called tangential acceleration. The second term, $\vec\omega\times\vec v$ is radially inwards and is called centripetal acceleration.

As you stated, the angular acceleration, tangential linear acceleration and distance between the reference point and the object are related using the following formula:

$$\vec{a} = \vec{\alpha} \times \vec{r}$$

$\vec{r}$ is merely the displacement vector between your choice of reference point and the object. The object needn't necessarily move in a circle for the formula to work.

The choice of reference point is arbitrary; you can choose any point. We often use the centre of mass or the centre of rotation as it simplifies the math but there is no rule which states that you should do your calculations about that point only.

There are two fundamental equation for linear acceleration and angular acceleration; these are:

$m \ddot{x} = F, \theta \ddot{\alpha} = M$.

it holds further $M = (x-x_0) \times F$ for some center of mass coordinate $x_0$. If you combine these equations you will get a relation between linear and angular acceleration.

In the case of a yoyo you must add some kinematic conditions like

$z = R \phi$

with the yoyo radius $R$, the height coordinate $z$ and the rotation angle $\phi$.