So in this system, and as well as for the stationary system of the observer, the acceleration of the car is given by:

$$ \boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta $$

The term $2 \dot{r} \dot{\theta} \textbf{e}_\theta $ is the Coriolis acceleration.

The car's acceleration in the rotating system and stationary system are not the same. The terms involving $\dot r$ and $\ddot r$ vanish if the car is going in a circle and the rotating observer is at the center of the circle. The acceleration in this case is identically zero in the rotating system but is nonzero in the stationary system.

The stationary observer only needs to know the frictional forces at the wheels of the car, the aerodynamic drag on the body of the car, and the acceleration from the car's engine to explain the car's motion. The stationary observer sees no Coriolis effect. That's a fictitious effect needed only by the rotating observer, and only if the rotating observer wants to use Newton's second law to explain the car's motion. This is one of several fictitious forces that arise in non-inertial frames.

What perplexes is me is that we are using a rotating and non-inertial system, i.e. the cylindrical coordinate system, and we make calculations in it that happen to satisfactorily describe the acceleration as the stationary observer's non-rotating and inertial system would measure them. What!?

There's no magic here. Those fictitious forces were specifically defined in a manner that allows non-inertial observers to describe motion via Newton's second law.

**Update: Using Newton's Second Law in a Non-Inertial Frame**

Suppose an inertial observer (I'm ignoring the rotation of the Earth) is in an observation stand, situated right above the circular track. The inertial observer knows about the individual forces acting on the car (wheel friction from turning the steering wheel, aerodynamic drag, force from engine torque (which also acts through the wheels)), sums these forces vectorially, and uses $\mathbf F = m\mathbf a$ to find the acceleration of the car.

Suppose another observer at the center of the track rotates so that the car appears to be stationary. Although there is a net horizontal force on the car, the car's acceleration from the rotating observer's perspective is zero (the car is stationary). Even worse, the rotating observer sees the inertial observer as circling around the track, opposite the rotating observer's rotation. Obviously a naive application of $\mathbf F = m\mathbf a$ doesn't work for the rotating observer. Newton's second law can be made to work by adding some fictional forces.

Let' make the track oval instead of circular. Now the rotating observer does see some acceleration. The car gets closer to and further from the observer as the car goes around the track. The acceleration as observed by the rotating observer is $\mathbf a_\text{rotating} = \ddot r \mathbf e_r$. The acceleration of the car as observed by the inertial observer, transformed to the rotating observers reference frame, is $\mathbf a_\text{inertial} = (\ddot r - r\dot\theta^2)\mathbf e_r + (r\ddot\theta + 2\dot r \dot\theta)\mathbf e_\theta$. (Note that this is the same expression in the opening question). The very first term, $\ddot r \mathbf e_r$, is the acceleration observed by the rotating observer. Using Newton's second law, this can be rewritten as
$$\frac {\mathbf F_\text{ext}} m = \mathbf a_\text{rotating} - r\dot\theta^2\mathbf e_r + (r\ddot\theta + 2\dot r \dot\theta)\mathbf e_\theta$$
or
$$\mathbf F_\text{ext} + mr\dot\theta^2\mathbf e_r - mr\ddot\theta\mathbf e_\theta -
2m\dot r \dot\theta)\mathbf e_\theta = m\mathbf a_\text{rotating}$$
Denote
$$\begin{aligned}
&\mathbf F_\text{centrifugal} && \equiv mr\dot\theta^2\mathbf e_r \\
&\mathbf F_\text{coriolis} && \equiv -2m\dot r \dot\theta)\mathbf e_\theta \\
&\mathbf F_\text{euler} && \equiv -mr\ddot\theta\mathbf e_\theta \\
&\mathbf F_\text{tot} && \equiv \mathbf F_\text{ext} + \mathbf F_\text{centrifugal} + \mathbf F_\text{coriolis} + \mathbf F_\text{euler}
\end{aligned}$$

With this, the expression that relates force and acceleration observed by the rotating observer simplifies to
$$\mathbf F_\text{tot} = m\mathbf a_\text{rotating}$$
Newton's second law!

## Best Answer

For how to do this mathematically ina simple way, see here, where they use \theta instead of \phi. This might also help. So, assuming that your $\phi$ coordinate is the azimuthal angle in the x-y plane, then the transformation to find the directional vector is

$$ \vec{e}_{\phi} = \frac{\partial \vec{r}}{\partial \phi} = -rsin(\phi)\hat{e}_{x} + rcos(\phi)\hat{e}_{y}$$

The unit vector is defined as, since the directional vectors are not necessarily of unit length,

$$ \hat{e}_{\phi} = \frac{\vec{e}_{\phi}}{|\vec{e}_{\phi}|}$$

So we have that,

$$ \hat{e}_{\phi} = \frac{\vec{e}_{\phi}}{r}$$

Next, to remove the explicit $\phi$ and $r$ dependence, we apply the coordinate transformation equations given here:

$$ r = \sqrt{x^{2} + y^{2}}$$

$$ \phi = arctan\Big(\frac{y}{x}\Big)$$

So, we have,

$$ \hat{e}_{\phi} = \frac{\vec{e}_{\phi}}{\sqrt{(rcos(\phi))^{2} + (rsin(\phi))^{2}}} = \frac{\vec{e}_{\phi}}{r}$$

$$ \hat{e}_{\phi} = \frac{-rsin(\phi)\hat{e}_{x} + rcos(\phi)\hat{e}_{y}}{r} = \frac{-y\vec{e}_{x} + x\vec{e}_{y}}{\sqrt{x^{2} + y^{2}}}$$

where we used the fact that $x = rcos(\phi)$ and $y = rsin(\phi)$.