I'm looking for a function on $R$ which is Lipschitz near a point $x_0$ but not differentiable at $x_0$, with the additional constraint that $f(x_0 + h) + f (x_0 - h) - 2f(x) = o(|h|)$ as $h \rightarrow 0$.
$xsin(1/x)$ (defined to be $0$ at origin) is not Lipschitz in any interval containing $0$ and it isn't differentiable at $0$, whereas $x^2sin(1/x)$ is differentiable at $0$ and Lipschitz. On the other hand, $|x|$ does not satisfy the "smoothness" condition.
Define $$f\ \colon\ x\longmapsto\begin{cases} x \cos\big(\ln(|x|)\big)&\text{if $x \neq 0$}\\ 0&\text{if $x=0$} \end{cases}$$
Then it is easy to check that $f$ is continuous on $\mathbb{R}$, and moreover $f$ is differentiable on $\mathbb{R}_+^*$, with $f'(x) = \cos \big( \ln(x) \big) - \sin \big( \ln (x) \big)$. So $f$ is Lipschitz continuous of order $2$ on $\mathbb{R}^+$. Similarly, you can prove that $f$ is differentiable on $\mathbb{R}_-^*$, and also Lipschitz continuous of order $2$ on $\mathbb{R}^-$. Hence $\ \ \ $ $f$ is Lipschitz continuous of order 2 on $\mathbb{R}$.
Moreover, $f$ is odd : $\forall y \in \mathbb{R},\ f(-y)=-f(y)$, so $\forall h \in \mathbb{R},\ f(0+h)+f(0-h)-2f(0)=0$, and thus $f$ satisfies your additional constraint at 0.
Finally, if we define the sequence $(x_n)=(e^{-n \pi})_{n \ge 0}$, we have for all $n \in \mathbb{N}$, $$\frac{f(x_n)-f(0)}{x_n-0}=\frac{x_n \cos(-n \pi)-0}{x_n-0} = \cos (n \pi) = (-1)^n$$
As $(x_n) \underset{n \to +\infty}{\longrightarrow} 0$ and $\big( (-1)^n \big)_{n \ge 0}$ diverges, we can conclude that $f$ is not differentiable at 0.
Here is the graph of the function : 