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$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth function such that $K(0)=1$ and $K(x)=0$ if $|x|\ge1$. Note that the intervals $2^{-n}[3/4,5/4]$ are disjoint for distinct integers $n$.

So, $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even continuous at $0$, because $f(2^{-k})=1\ne f(0)$ for all natural $k$.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth function such that $K(0)=1$ and $K(x)=0$ if $|x|\ge1$. Note that the intervals $2^{-n}[3/4,5/4]$ are disjoint for distinct integers $n$.

So, $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=2f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even differentiable at $0$, because $f(2^{-k})=2^{-k}$ and $f(\frac34\,2^{-k})=0$ for all natural $k$.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth nonzero function such that $K(x)=0$ if $|x|\ge1$.

Then $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even continuous at $0$.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth function such that $K(0)=1$ and $K(x)=0$ if $|x|\ge1$. Note that the intervals $2^{-n}[3/4,5/4]$ are disjoint for distinct integers $n$.

So, $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even continuous at $0$, because $f(2^{-k})=1\ne f(0)$ for all natural $k$.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a nonzero smooth function such that $K(x)=0$ if $|x|\ge1$.

Then $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=f(x)$ for all real $x$. Then you condition holds, but $f$ is not even continuous at $0$.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth nonzero function such that $K(x)=0$ if $|x|\ge1$.

Then $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even continuous at $0$.