Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$.
In this case, the angle, $\theta$, is the equivalent of distance traveled in linear kinematics, and it has units of radians (${rad}$). (Radians, being unit less, are to some extent a place holder, but place holders can be very useful, so keep them in mind.) So, then the rate of change of angle with respect to time, $\omega$, has the units of ${rad}/s$. Angular acceleration, $\alpha$, will then have units of ${rad}/s^2$.
With those in mind, you can immediately tell that $a_c = \frac{v^2}{r}$ is not an angular acceleration, but a linear acceleration, as described by Peter. Similarly, angular acceleration is not directly related to force, but to torque, $\tau = I \alpha$, where $I$ is the moment of inertia. (From a mathematical perspective, the moment of inertia is the second moment of the mass distribution where the center of mass is the first moment.) Torque has the units ${kg}\ m^2/s^2$, where the radians were dropped. Note, it has units of energy, or $(Force)(distance)$, and $\tau = r \times F$.
On any single parameter curve in $\mathbb{R}^n$, $n\geq2$, the derivative with respect to that parameter always lies tangent to the curve. The derivative is literally showing us how the position is going to change. From a physics perspective, you can think of this as attaching the velocity and acceleration vectors to the moving object, itself, as in drawing a free body diagram.
To be concrete, for uniform circular motion, the position is
$$r(t) = R( \cos(t) \hat{i} + \sin(t) \hat{j} )$$
where $R$ is the radius of the circle, $\hat{i}$ and $\hat{j}$ are the unit vectors in the $x$ and $y$ directions, respectively, and the velocity is
$$v(t) = R( -\sin(t) \hat{i} + \cos(t) \hat{j} ).$$
Note, that the velocity is perpendicular to the position which is a property of circular motion. From this, you should be able to mathematically demonstrate that the acceleration is perpendicular to the velocity and anti-parallel to the position. I'll leave you with the problem of understanding why this makes sense physically, also.