[https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec06.pdf], on page 2, equation #4, it is stated that the acceleration is the derivative of the rate of change in the position vector $r$,
$$a = \frac{d}{dt} \left(\frac{dr}{dt}\right) = \frac{d}{dt} \left( \frac{dr}{ds} \cdot s' \right) = \left(\frac{d^2r}{ds^2}\cdot \left(s'\right)^2 \right)+ \left(s''\cdot \frac{dr}{ds} \right).$$
The problem is with the $\left(s'\right)^2$ part isn't it supposed to be $s'$ only from the product rule?
We need the chain rule (CR) and product rule (PR). Since $\frac{dr}{dt}=\frac{dr}{ds}s^\prime$, $$\begin{align}\frac{d^2r}{dt^2}&\stackrel{\text{CR}}{=}\underbrace{s^\prime\frac{d}{ds}}_{\frac{d}{dt}}\left(\frac{dr}{ds}s^\prime\right)\\&\stackrel{\text{PR}}{=}s^\prime\left(\frac{d^2r}{ds^2}s^\prime+\frac{dr}{ds}\underbrace{\frac{ds^\prime}{ds}}_{s^{\prime\prime}/s^\prime}\right)\\&=s^{\prime2}\frac{d^2r}{ds^2}+s^{\prime\prime}\frac{dr}{ds}.\end{align}$$