We say that a function $f:\mathbb{R}^n\to\mathbb{R}^m$ is point Lipschitz at $x_0\in\mathbb{R}^n$ if there exists a neighborhood $U$ of $x_0$ and a constant $L>0$ such that: \begin{align*} ||f(x)-f(x_0)|| \leq L ||x-x_0|| \quad \forall x\in U. \end{align*} Moreover, we say that a function is locally point Lipschitz at $x_0$ if there exists a neighborhood $U_2$ of $x_0$ such that $f$ is locally point Lipschitz at all points of $U_2$. Alternatively, $f$ is locally Lipschitz at $x_0\in\mathbb{R}^n$ if there exists a neighborhood $U_3$ of $x_0$ and a constant $L_3>0$ such that: \begin{align*} ||f(x)-f(y)|| \leq L_3 ||x-y|| \quad \forall x,y \in U_3. \end{align*} Clearly, all locally Lipschitz functions at $x_0$ are locally point Lipschitz at $x_0$. Moreover, there are functions (see this post) that are point Lipschitz at a point but not locally Lipschitz at it. However, all such functions seem to be point Lipschitz but not locally Lipschitz only at isolated points.
Hence my question: If $f$ is locally point Lipschitz at $x_0$, does there always exist a neighborhood $V$ of $x_0$ such that $f$ is locally Lipschitz for all points of $V\backslash \{x_0\}$? Is there a counterexample for this statement?
No, such a neighborhood need not exist. Here is a counterexample.
Let $f\colon \mathbb R^n\to \mathbb R$ be a function that is locally point Lipschitz at $0$, not locally Lipschitz at $0$, and locally Lipschitz everywhere else, as you have described. By multiplying by a bump function, we may assume $f$ is supported on the unit ball $B$ centered at $0$. With no loss of generality we may assume $f\leq 1$ everywhere on $B$.
Now let $x_n\to x_0$ be a sequence of distinct points converging to $x_0$, and observe this means there are closed balls $B_n$ of radius $r_n$ centered at $x_n$ such that the balls $2B_n$ are pairwise disjoint and do not intersect $x_0$. For each $n$ let $g_n(x)= \|x_n-x_0\|f(\frac{x-x_n}{r_n})$, and observe $g_n$ is supported on $B_n$ and bounded by $\|x_n-x_0\|$ on $B_n$. Let $g=\sum_n g_n$.
Then $g$ is certainly point-Lipshitz everywhere - this follows easily from the definition everywhere but $x_0$, and at $x_0$, for any $x\in B_n$ we have $g(x)=g_n(x)$, so $$|g(x)-g(x_0)|=|g_n(x)-0|\leq \|x-x_n\|\leq \|x-x_0\|\text{,}$$ and at any $x\notin \bigcup_n B_n$, $g(x)=0$.
On the other hand, since $g$ restricted to each $B_n$ coincides with a scaled and translated version of $f$, $g$ is not locally Lipschitz at any point $x_n$. Since every neighborhood of $x_0$ contains infinitely many $x_n$, the counterexample is established.