Let $f \in L^2(0,2\pi)$ be taken such that $f$ and $f'$ are absolutely continuous on $[0,2\pi]$ with $f(0) = f(2\pi)$ and $f'(0)= f'(2\pi).$ Is this sufficient to conclude from this that $f'' \in L^2(0,2\pi)$? We clearly have $f'' \in L^1(0,2\pi)$ because $f'$ is absolutely continuous, but I currently don't see that $f'' \in L^2(0,2\pi)$ follows. I suspect that it does not hold, but I don't really know.
If anything is unclear, please let me know.
Nope, interpolation inequalities work in the other way: provided that $f,f''\in L^2$, then $f'\in L^2$, but in your case you may consider
$$ f(x) = \sin\left(\frac{x}{2}\right)^\frac{3}{2} $$ and check that $f,f'\in L^2(0,2\pi)$ while $f''$ does not belong to $L^2(0,2\pi)$.
Ok, $f'(0)=f'(2\pi)$ is not fulfilled, but it is easy to fix by considering: $$ g(x) = \left|\sin x\,\right|^{\frac{3}{2}}\cdot\operatorname{sign}(\sin x).$$