Let $B$ be a compact convex set on the complex plane, containing zero in its interior. The boundary $\partial B$ of $B$ has the polar parametrization $\mathbf p:\mathbb R\to \partial B$ assigning to each real number $t$ the unique point of the intersection of $\partial B$ with the ray $\{re^{it}:r>0\}$. The function $\mathbf p$ is $2\pi$periodic and has two main properties of convex functions: it has twosided derivatives $\mathbf p'_(t)$ and $\mathbf p'_+(t)$ for every $t\in \mathbb R$ and those twosided derivatives have bounded variation on any bounded subset of the real line. Consequently $\mathbf p$ is twice differentiable almost everywhere. Those facts are moreorless "obvious" but I do not see short proofs (or even short reduction to the known properties of convex functions). Maybe this is alredy done somewhere? Just to insert a reference and not spend time for reproving known results. Please, help.
If $f(x)=\min\{s>0:x/s\in B\}$ is Minkowski functional of $B$, then $f$ is a convex function on the plane and ${\bf p}(t)=\frac{e^{it}}{f(e^{it})}$. I think your claims now follow from the properties of convex functions. For example, you may locally choose a smooth function $h(t)\in (0,\infty)$ such that $h(t)e^{it}$ runs over a segment $[AB]$ on the plane when $t$ runs over a small enough interval $\Delta$, then write ${\bf p}(t)=\frac{h(t)e^{it}}{f(h(t)e^{it})}$, and choose the parameter on $[AB]\ni h(t)e^{it}$ as a new variable.

$\begingroup$ Thank you very much for the suggestion. I tried to use the convexity of the ball in an affine system and had problems with the proof of the boundedness of variation of the (onesided) derivatives of $\mathbb p$. With your "Minkowski" approach this problem does not arize. So, thank you once more. $\endgroup$ – Taras Banakh Oct 30 '19 at 9:27