Both expressions for the acceleration are correct, the problem is that they do not refer to the same thing.
In any kind of motion (of a point particle) you should know that the velocity is always tangent to the trajectory of the object. The case of circular motion is the same.
You may not know this but the right way to express angular velocity is as a vector (just like with regular velocity) that points in the direction of the axis of rotation. The magnitude of the vector gives the instantaneous rate of rotation.
Through a vector relation, we get the vector velocity due to the rotation:
$$\overrightarrow{\omega}\times\overrightarrow{r}=\overrightarrow{v}$$
In the case of circular motion $\overrightarrow{\omega}$ and $\overrightarrow{r}$ are perpendicular, so that
$$\left|\overrightarrow{\omega}\times\overrightarrow{r}\right|=\left|\overrightarrow{\omega}\right|\left|\overrightarrow{r}\right|\mathrm{sin}(\phi)=\omega r\text{ }\mathrm{sin}(\phi)=\omega r$$
That is the expression you posted in the beginning, but note that it is only the magnitude of the velocity vector. This is important because the acceleration is also a vector that gives you both the magnitude and the direction of change of the velocity.
To get the acceleration you simply differentiate the velocity with respect to time (as always):
$$\overrightarrow{a}=\frac{d}{dt}\left(\overrightarrow{v}\right)=\frac{d}{dt}\left(\overrightarrow{\omega}\times\overrightarrow{r}\right)$$
And you use the fact that the cross product follows Leibniz's rule (careful not to change the order of the products):
$$\overrightarrow{a}=\frac{d\overrightarrow{\omega}}{dt}\times\overrightarrow{r}+\overrightarrow{\omega}\times\frac{d\overrightarrow{r}}{dt}=\boxed{\overrightarrow{\alpha}\times\overrightarrow{r}+\overrightarrow{\omega}\times\overrightarrow{v}}$$
In circular motion, $\overrightarrow{\alpha}$ goes in the same direction as $\overrightarrow{\omega}$ because the axis of rotation doesn't change. The magnitude of the first term is then simply $\alpha r$, your first expression for acceleration. The second term is also simply $\omega v$ but by using $v=\omega r$, the magnitude yields $v^2/r$, your second expression.
So both are valid, but if you look carefully you should realize that the $\overrightarrow{\alpha}\times\overrightarrow{r}$ vector points in the direction of the velocity, it is thus called the tangential acceleration $\overrightarrow{a_{t}}$. On the other hand, the $\overrightarrow{\omega}\times\overrightarrow{v}$ points towards the center of the circle, and is perpendicular to $\overrightarrow{a_{t}}$, it is therefore called the normal acceleration $\overrightarrow{a_{n}}$.
Just a little extra: You can see that the tangential acceleration is generated by the angular acceleration, this simply means that if you spin your particle faster the velocity increases. But the normal acceleration is always there (as long as $\omega\neq 0$) this is because the velocity is always changing its direction along the circular trajectory. The normal acceleration makes it change so that it is always tangent to this circle.