I need to come from $r''=\frac{d^2r}{ds^2}$ to $\ddot{r}=\frac{d^2r}{dt^2}.$ I know that $r' = \frac{\dot{r}}{|\dot{r}|}.$ And more $1/\dot{s} = t' = \frac{1}{\sqrt{\dot{r}\cdot\dot{r}}}.$ You can see even more here
so i try this way $$r''=\frac{d^2r}{ds^2}=\frac{d}{ds}\bigg(\frac{dr}{ds}\bigg)=\frac{d}{ds}\bigg(\frac{\dot{r}}{\sqrt{\dot{r}\cdot\dot{r}}}\bigg)=\frac{\frac{d^2r}{dt\ ds}\sqrt{\dot{r}\cdot\dot{r}}-\dot{r}(\sqrt{\dot{r}\cdot\dot{r}})'}{\dot{r}\cdot\dot{r}}=\frac{\frac{\ddot{r}}{\sqrt{\dot{r}\cdot\dot{r}}}\sqrt{\dot{r}\cdot\dot{r}}-\dot{r}\frac{2\frac{\ddot{r}}{\sqrt{\dot{r}\cdot\dot{r}}}\cdot\dot{r}}{2\sqrt{\dot{r}\cdot\dot{r}}}}{\dot{r}\cdot\dot{r}}=\frac{\ddot{r}-\dot{r}\frac{\ddot{r}\cdot\dot{r}}{\dot{r}\cdot\dot{r}}}{\dot{r}\cdot\dot{r}}.$$
It looks for me like i have made a mistake. I also need it for $r'''$ to $\overset{...}{r}.$ But may be i can manage it myself later. Thanks a lot.
Your " from $r''$ to $\ddot r\>$" is somewhat misleading. What you want is the following:
You are given a function $t\mapsto r(t)$ which then determines a certain function $\sigma:\>t\mapsto s(t)$ whereby $$\dot s(t)=|\dot r(t)|>0\tag{1}$$ throughout. It follows that $\sigma$ has an inverse function $\sigma^{-1}: \> s\mapsto t(s)$, so that we can "express $r(\cdot)$ in terms of $s\>$", which means that we are interested in the pullback $$\bar r(s):=r\bigl(t(s)\bigr)$$ We are now interested in the successive derivatives of $\bar r(\cdot)$ with respect to $s$, which we shall denote by $r'$, and so on.
Now given any function $u:\>t\mapsto u(t)$ we can form the pullback $$\bar u(s):=u\bigl(t(s)\bigr)\ ,$$ and the chain rule then gives $$\bar u'(s)=\dot u\bigl(t(s)\bigr)\> t'(s)={1\over\dot s\bigl(t(s)\bigr)}\dot u\bigl(t(s)\bigr)\ .$$ This can be condensed to the rule $$\bar u'={\dot u\over\dot s}\ .\tag{2}$$ Choosing $u:=r$ we immediately obtain $$\bar r'={\dot r\over\dot s}\ .\tag{3}$$ In order to compute $\bar r''$ we put $u:={\dot r\over|\dot r|}$. Applying the rule $(2)$ to $(3)$ we then obtain $$\bar r''=\bar u'={1\over\dot s}\left({\dot r\over\dot s}\right)^\cdot={\dot s\ddot r-\ddot s\dot r\over\dot s^3}\ .\tag{4}$$ It becomes apparent that we have to compute $\ddot s$ in terms of $r$. From $(1)$ we obtain $$\ddot s={d\over dt}\sqrt{\dot r\cdot\dot r}={\ddot r\cdot \dot r+\dot r\cdot\ddot r\over 2\sqrt{\dot r\cdot\dot r}}={\dot r\cdot\ddot r\over\dot s}\ .$$ Plugging this into $(4)$ we finally obtain your result $$\bar r''={\dot s^2 \ddot r-(\dot r\cdot\ddot r )\dot r\over\dot s^4}\ .$$