Linear acceleration vs angular acceleration equation
You made a mistake in assuming that the angular acceleration ($\alpha$) is equal to $v^2/r$ which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle $\theta$. This is very similar to how the linear acceleration is defined.
$$a=\frac{d^2x}{dt^2} \rightarrow \alpha=\frac{d^2\theta}{dt^2}$$
Like the linear acceleration is $F/m$, the angular acceleration is indeed $\tau/I$, $\tau$ being the torque and I being moment of inertia (equivalent to mass).
I also am confused on what exactly 'V' (tangential velocity) represents and how it's used. Is it a vector who's magnitude is equal to the number of radians any point on a polygon should rotate?
The tangential velocity in case of a body moving with constant speed in a circle is same as its ordinary speed. The name comes from the fact that this speed is along the tangent to the circle (the path of motion for the body). Its magnitude is equal to the rate at which it moves along the circle. Geometrically you can show that $v = r\omega$.
$a_c = \frac{v^2}{r}$ isn't angular acceleration. It's the magnitude of the linear acceleration towards the centre of an object following a circular path at constant angular velocity. Angular acceleration is the derivative of angular velocity, and the analogue of Newton's second law is that angular acceleration equals torque divided by moment of inertia.
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.