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.

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.

A little bit more analysis:

Let $D_{h} f(x)$ be the difference quotient of the function $f$ $$ D_h f(x) = \frac1h (f(x+h) - f(x)) $$ The expression you wrote down is $$ \underbrace{D_h D_h D_h \cdots D_h}_{n \text{ times}} f(x) $$

$n$-times differentiability requires the limit $$ \lim_{h_n \to 0} \lim_{h_{n-1}\to 0} \cdots \lim_{h_1\to 0} D_{h_n} \cdots D_{h_1} f(x) $$ to exist.

Your condition only requires the limit to exist along the diagonal, so can be quite far from enough.