How do I know what variable to use for the chain rule? In my textbook the tangential acceleration is given like this:
$$a_t=\frac{dv}{dt}=r\frac{dw}{dt}$$
$$a_t=rα$$
I understand that the chain rule is applied here like this:
$$a_t=\frac{dv}{dt}=\frac{dv}{dw}\frac{dw}{dt}=rα$$
What I don't understand is why we have to apply the rule in this specific way. Say I write like this:
$$a_t=\frac{dv}{dθ}\frac{dθ}{dt}$$
This way, I end up with entirely different result. How do I know how the chain rule must be applied?
 A: 
What I don't understand is why we have to apply the rule in this specific way?
  How do I know how the chain rule must be applied?

We don't have to. You don't know. Somebody just found out that by using that specific method, the result ended up neat and simple.
Nothing is wrong with another method. You get the same thing in another expression. Let's try to interpret the terms in your result:
$$a_t=\frac{dv}{dθ}\frac{dθ}{dt}$$
$\frac{dθ}{dt}$ clealy equals $\omega$. $\frac{dv}{dθ}$ is a bit tougher - something like instantanous speed change per angle change. If you have a way to measure this, then the formula:
$$a_t=\frac{dv}{dθ}\omega$$
is just as useable. Just not as neat. I mean, it is kinda smart that $a_t=r\alpha$ has the same shape as $v=r\omega$ and $s=r\theta$. Makes the overview much better, when we can end with a similar and simple result.

Note, if you already have the expression $v=r\omega$ at hand, then you don't even need chain rules to reach the simple expression:
$$a_t=\frac{dv}{dt}=\frac{d(r\omega)}{dt}=r\frac{d\omega}{dt}=rα$$
A: In general you are allowed to use any parameter $q$ to describe the motion of an object as $\vec{r}(q)$ where the parameter changes with time $q=q(t)$. The parameter can be an angle or a distance or any combination that best makes sense.
So now you have expressions for velocity and acceleration defined from the chain rule
$$ \vec{v} = \frac{{\rm d} \vec{r}}{{\rm d} t} 
= \frac{{\rm d} \vec{r}}{{\rm d} q} \frac{{\rm d} q}{{\rm d}t} = \vec{r}\,' \dot{q} $$
$$ \vec{a} = \frac{{\rm d} \vec{v}}{{\rm d} t} 
= \frac{{\rm d} \vec{v}}{{\rm d} q} \dot{q} + \frac{{\rm d} \vec{v}}{{\rm d} \dot{q}} \ddot{q} = \frac{{\rm d} (\vec{r}\,' \dot{q})}{{\rm d} q} \dot{q} + \frac{{\rm d} (\vec{r}\,' \dot{q})}{{\rm d} \dot{q}} \ddot{q} =\\ 
\vec{a} =\vec{r}\,'' \dot{q}^2 + \vec{r}\,' \ddot{q}$$
So what the book did is set $q=\theta$, $\dot{q}= \omega$ and $\ddot{q} = \alpha$. This results in the following for circular motion
$$\vec{r} = R\,\vec{n}(\theta)$$ where $\vec{n}$ is the normal direction away from the center (which varies by $\theta$). Now you have
$$ \vec{v} = \frac{{\rm d}R \vec{n}}{{\rm d}\theta} \omega = R \omega \frac{{\rm d}\vec{n}}{{\rm d}\theta} = R \omega \vec{e}$$
where $\vec{e}$ is the tangetial direction
$$\vec{a} = \frac{{\rm d}R \omega \vec{e}}{{\rm d}\theta} \omega +\frac{{\rm d}R \omega \vec{e}}{{\rm d}\omega} \alpha = R\omega^2\frac{{\rm d}  \vec{e}}{{\rm d}\theta}  +R \alpha \vec{e}=R \alpha \vec{e} - R\omega^2 \vec{n}$$
So the tangential component is $R \omega$ and the centripetal $R \omega^2$.
A: In the case of a circular motion, the tangential velocity $v$ can be expressed in terms of $\omega$ and $r$, i.e. $ v= \omega r$.
The formula in your textbook $$a=r\frac{d\omega}{dt} $$is obtained by taking the derivative of the previous equation, while taking into account that $r$ is constant.
Of course, you could consider $v$ as a function of $\theta$, i.e. $v=v(\theta)$. However, there is no simple, generally applicable rule like the one above relating $v$ and $\theta$. So, the formula $$a=\omega\frac{dv}{d\theta}$$ is correct but does not help us a lot.
In case of a circular motion with uniform acceleration, we get $$\theta = \frac{1}{2} \alpha t^2$$ or $$v=r \sqrt{2\alpha \theta}$$
Applying now your alternative method (derivative to $\theta$), leads to $a=r\alpha$, just like your textbook formula does.
