To avoid confusion, I shall denote $f=f(r)$ and $F(x)=f(|x|)$ the corresponding radial function.

Because of Van der Corput-Schaake Inequality, it is enough to prove the following more general inequality: for every point $x\ne0$, integer $k\ge1$ and unit vector $e$,
$$|D^kF_xe^{\otimes k}|\le\sup_{0\le s\le r}|f^{(k)}(s)|\quad ?$$
This is obvious for $k=1$. Let me first prove that it is true for $k=2,3$. Then I'll give a general strategy for a proof. In the following, I denote
$$\nu:=\frac{x\cdot e}r\in[-1,1].$$

For $k=2$, one has
$$D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1rf'+\nu^2f''.$$
Because $F$ is smooth, we know that the derivatives of $f$ of odd order vanish at $0$. Using $f'(0)=0$, we obtain
$$D^2F_xe^{\otimes 2}=(1-\nu^2)\frac1r\int_0^rf''(s)ds+\nu^2f''=\langle\mu_2,f''\rangle.$$
There remains to check that the measure $\mu_2$ has total mass $1$ (*i.e.* is a probability). This is obvious.

If $k=3$, we have
$$D^3F_xe^{\otimes 3}=3(\nu-\nu^3)\left(\frac1rf''-\frac1{r^2}f'\right)+\nu^3f'''.$$
Using again $f'(0)=0$, we obtain
$$D^3F_xe^{\otimes 3}=\langle\mu_3,f'''\rangle$$
where
$$\mu_3:=\nu^3\delta_{s=r}+3(\nu-\nu^3)\frac{s}{r^2}\chi_{[0,r]}(s).$$
Again, one checks that
$$|\mu_3|=|\nu^3+\frac32(\nu-\nu^3)|\le1.$$

The general strategy is to prove that if $r=|x|$, then
$$D^kF_xe^{\otimes k}=\langle\mu_k,f^{(k)}\rangle$$
for some measure $\mu_k$ over $[0,r]$. Mind that the coefficients of $\mu_k$ involve only $\nu$ and $r$. The measure $\mu_k$ can be calculated from the Faa di Bruno Formula (I suspect that $\mu_k$ has a constant sign). To conclude, one has to check that its total mass is less than $1$. This total mass involves only $\nu$ ; it is likely to be the absolute value of a polynomial $P_k(\nu)$.

More precisely, $\mu_k$ is the sum of the Dirac mass $\nu^k\delta_{x=r}$ and a continuous density over $s\in(0,r)$ :
$$\frac1rN_k(\nu,t),\qquad t:=\frac sr.$$
The density is determined by induction with the following rules
$$\partial_t(N_{k+1}-\nu tN_k) = -(1-\nu^2)\partial_\nu N_k, $$
$$N_{k+1}(\nu,1) = \nu N_k(\nu,1)+k\nu^{k-1}(1-\nu^2),$$
and we have $N_k(\nu,0)$ if $k$ is odd. The second line gives explicitly
$$N_k(\nu,1)=\frac{k(k-1}2\nu^{k-2}(1-\nu^2).$$ The first few $N_k$'s are
$$N_1\equiv0,\quad N_2=1-\nu^2,\quad N_3=3\nu(1-\nu^2)t,$$
$$N_4=\frac32(1-\nu^2)(5\nu^2-1)(t^2-1)+6\nu^2(1-\nu^2),$$
$$N_5=\frac32\nu(1-\nu^2)(5\nu^2-1)t(t^2-1)+6\nu^3(1-\nu^2)(t-1)-2(3\nu-5\nu^3)(1-\nu^2)(t^3-3t+2)-12\nu(1-\nu^2)(1-2\nu^2)(t-1)+10\nu^3(1-\nu^2).$$
One verifies easily that $\mu_4$ is a probability (Question : is $\mu_{2\ell}$ always a probability ?).

I checked that for every value of the angle $\nu$,
$$0\le\langle\mu_5,{\bf1}\rangle\le1.$$
However, I gave up about the sign of $N_5$, and therefore I cannot claim that the total mass $|\mu_5|\le1$.