If a function is non-continuously differentiable at a point $c$, then it has a derivative at $c$, but the derivative is not continuous at $c$. This makes no sense to me. It is seemingly the conjunction of two contradictory statements:
$$\lim_{x \rightarrow c} \frac{f(x) - f(c)}{x-c} \in \Bbb R \tag1$$
$$$$
$$\lim_{x \rightarrow c}\lim_{x \rightarrow c} \frac{f(x) - f(c)}{x-c} \not\in \Bbb R \tag2$$
The second statement is true, because $f'(x)$ is discontinuous at $c$. Thus, either the left- and right-hand limits are not equal, or, the limit does not exist at all, both of which are captured by $(2)$. But composing limits is idempotent, so $(1)$ and $(2)$ are contradicting each other.
What am I missing?
The problem is in your second statement. Indeed, as you've written it, the first and second equations contradict each other. To calculate the derivative at a point $c$, we take a limit of a ratio involving $f$ as a dummy variable (which you write as $x$) approaches $c$. This dummy variable only exists within the context of this limit. To check the continuity of of the derivative at a point $c$, we need to calculate the limit $f'$ as a dummy variable approaches $c$--in particular this needs to be a new dummy variable. So I'd rewrite your second equation as
$$f''(c) = (f')'(c) = \lim_{x \to c} f'(x) = \lim_{x \to c} \lim_{y \to x} \frac{f(x) - f(y)}{x - y} \; \text{does not exist}.$$
It may help to see how discontinuities happen in derivatives by looking at some real examples: for that, take a look at this pdf which works through the classic example $$f(x) = \begin{cases} x^2\sin(1/x) & x \neq 0 \\ 0 & x = 0\end{cases}$$ (or check it for yourself with just the limit definition!)