$\newcommand\de\delta$No. E.g., let 
$$f(x):=x^3\sin\frac1x$$
for $x\in(0,1/2]$, with $f(0):=0$. Then $f$ can be obviously extended to a continuously differentiable function $f$ on $\mathbb R$ supported on $[0,1]$. 

Moreover,  
$$rhs(\de):=\de^{-1}\sup_{0<h\le\de}|f(2h)-2f(h)+f(0)| \\ 
=\de^{-1}\sup_{0<h\le\de}\Big|-2 h^3 \Big(\sin\frac1h-4 \sin \frac{1}{2h}\Big)\Big|=O(\de^2)=o(\de), 
$$
while for $\de=\dfrac1{2(2n+1)\pi}$ and natural $n\to\infty$
$$lhs(\de):=\sup_{0<h\le\de}|f'(2h)-2f'(h)+f'(0)| \\ 
\ge |f'(2\de)-2f'(\de)+f'(0)| \\
=4 \de  \sin ^2\frac{1}{4 \de }\, 
\Big|6 \de  \sin
   (\frac{1}{2 \de })-2 \cos (\frac{1}{2 \de
   })-1\Big| \\ 
\sim4 \de  \sin ^2\frac{1}{4 \de }=4\de. 
$$
So, $lhs(\de)$ is not $O(rhs(\de))$.