Unfortunately, there are two ways to interpret an “angular acceleration” and a “radial acceleration.” You will have to ask your textbook or instructor which is meant.
This is easiest to see for the radial acceleration of uniform circular motion. I can give you two correct answers. One is that in uniform circular motion $r(t)$ is constant so $\ddot r=0$ and this is what I mean by radial acceleration, so the radial acceleration is zero. The other meaning would be, the component of the acceleration vector in the radial direction: this is not zero but $-v^2/r$ for uniform circular motion. Which answer is correct depends on exactly what you are interested in.
You ask if we can say that the centripetal acceleration is zero simply because something doesn't happen to be moving on a circular trajectory, and I would answer this in the negative. Instead, calculus is all about approximating things with other things, and you can approximate a curving line with a circle. Indeed we can define a radius of curvature as $r=v^2/|a|$ or so, something like that, even if we are not on a circular trajectory.
More generally if I write $$\mathbf r =\begin{bmatrix}x(t)\\y(t)\end{bmatrix}= r\hat r =r(t) \begin{bmatrix}\cos(\theta(t))\\\sin(\theta(t))\end{bmatrix},$$then the consequence is that the velocity is$$\mathbf v=\dot{\mathbf r}=\dot r\hat r + r\dot\theta\hat\theta$$ and then one more derivative gives$$\mathbf a=\ddot{\mathbf r}=(\ddot r-r\dot\theta^2)\hat r +(r\ddot\theta +2\dot r\dot\theta)\hat\theta.$$this gives the conversion between the two different definitions, and you can interpret these terms as being centrifugal and coriolis force in a co-moving frame if you would like.
