First, let's write out what your expression requires when $n = 2$:

$$ \lim_{h \to 0} \frac{f(x) + f(x + 2h) - 2 f(x+h)}{h^2} $$

is required to exist for all $x \in U$. 

Let $U = (-1,1)$. Take $f(x) = |x|$. 

When $x \neq 0$, there exists a sufficiently small interval $I_x$ around $x$ such that $f|_{I_x}$ is $C^2$, and clearly the expression evaluates to $0$. 

When $x = 0$, you have that for any $h$:

$$ \frac{f(0) + f(2h) - 2 f(h)}{h^2} = 0 $$

and hence the limit exists and equals 0. 

The absolute value function is clearly not twice differentiable at 0.