This may be trivial but I cannot fully wrap my head around it:
Suppose we have a simple closed regular differentiable plane curve $\alpha : \mathbb{R}\rightarrow \mathbb{R}^2$ parametrized in arc length. Moreover suppose that is it positively oriented (the normal vector always point to the interior of the curve) then define $\beta(s):=\alpha(-s)$ show that it is negatively oriented (and v.v. of course). My issue is that while taking derivatives I get lost in changing signs, thanks in advance for the help.
(1) If you define the normal as the derivative of $\frac{\alpha'(s)}{\|\alpha'(s)\|}$, as you suggest in your question, then the statement is not correct. The fact that the normal vector points inwards doesn't tell anything about how you walk through the curve (counterclockwise or clockwise).
Example: $$ \begin{align*} \alpha_1(s) &= \left( \cos (s),\sin(s)\right) \\ \alpha_2(s) &= \left( \cos (-s),\sin(-s)\right) \end{align*} $$
Taking the second derivatives gives the normal vectors. In both cases, the normal vectors point inwards. However, $\alpha_1$ parametrises the circle counterclockwise; $\alpha_2$ clockwise.
(2) However, for plane curves one often defines the normal as $N=JT$, where $J$ stands for "rotating the vector 90 degrees counterclockwise". In this case, the statement is true (just use the right hand rule).