This function for air resistance (drag) supposedly shows a Taylor expansion. I understand the basics of Taylor expansion, but can't see how I would get this answer. If someone could elaborate and explain the intermediate steps and how these last terms should be understood it would be very helpful.
Let f be a function. If we have that $f$ is differentiable at a point $a$, we have that $$f(x)=f(a)+f'(a)(x-a)+r(x)$$ With $$\lim\limits_{x \to a} \frac{r(x)}{x-a}=0$$ So around $a$, we have the best linear approximation if we neglect $r$: $$f(x) \approx f(a)+f'(a)(x-a)$$
In your problem, we have that $f$ is a function of the velocity $v$: $f(v)=Cv^2$. If we pick a velocity $v_a$, we have the following approximation around $v_0$: $$f(v) \approx f(v_a) + f'(v_a)(v-v_a)$$ $$Cv^2 \approx Cv_a^2+\left.\frac{\mathrm{d} (Cv^2)}{\mathrm{d}v}\right|_{v=v_a}(v-v_a)$$ So we have what you wanted.