$\newcommand\de\delta$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 $O(rhs(\de))$.