Local Lipschitz continuity: existence and uniqueness of a solution to a ODE Cauchy problem

Question

The version of the Picard–Lindelöf that I know, which is possibly one of the most common statements of the theorem states that, given a closed rectangle $D\subset \mathbb{R}\times\mathbb{R}^n$ and let $(x_0, y_0) \in \operatorname{int} D$ and $f:D\rightarrow\mathbb{R}$, then if $f$ is continuous and $f$ is Lipschitz continuous with respect to $y$, $\exists\varepsilon>0$ such that the Cauchy problem

the Cauchy problem $$y'=f(x,y), \qquad y(x_0)=y_0 \tag{CP}\label{CP}$$ admits a unique solution on the interval $[x_0-\varepsilon, x_0+\varepsilon]$ (cf. e.g. the wiki article I linked to above, from which I quoted this version of the theorem). Note that the solution is defined in a small intervall around the initial point.

Now, I wondered whether we could relax the hypothesis to local Lipschitz continuity in the second variable, since the final result is after all local (in the sense of the previous sentence). I don't think that should be a problem, as in the proof Lipschitz continuity only comes in using the Banach fixed point theorem and I guess that up to restrictions things should work the same.

In fact, I could find some results here on SE, such as this question, where they say the problem is that the existence of a global solution is not guaranteed, although locally a unique solution is determined. The Picard–Lindelöf theorem already states that the solution is local. It's not like the theorem is asserting the existence of a unique solution.$^{(a)}$ Then what's so important about Lipschitz continuity, why isn't the theorem stated requiring local Lipschitz continuity with respect to $y$? Why do the answers above do as if that only were the case with local Lipschitz continuity?

It's worth discussing why I'm wondering this. A notable example is that $f\in C^1$. If the domain if compact (as above), then $f$ is Lipschitz continuous. But, in other versions of the theorem, the domain is not (e.g. an open set), then only local Lipschitz continuity is implied. As an example, here I quote a (translated) theorem from Ref. 1

For all open sets $U\in\mathbb{R}^n$ and functions $X^1... X^n\in C^\infty (U)$ we have:

1. $\forall t_0\in\mathbb{R}, x_0\in U$, there exist $\delta>0$ and an open neighbourhood of $x_0$, $U_0\subseteq U$ such that $\forall x\in U_0$ there exists a curve $\sigma_x:(t_0-\delta, t_0+\delta)\rightarrow U$ which is solution of the Cauchy problem $$\sigma'^j(t)=X^j(\sigma(t)), \quad \sigma(t_0)=x\quad j=1,...n; \tag{CP2}\label{CP2}$$
2. The map $\Theta:(t_0-\delta, t_0+\delta)\times U_0\rightarrow U$ defined by $\Theta(t,x)=\sigma_x(t)$ is $C^\infty$;
3. Two solutions of \eqref{CP2} always coincide in the intersection of their domains.

I included this because it may give more context. The core question is boldfaced.

References

1. Geometria Differenziale, M. Abate, F. Tovena. Springer, 2011. Teorema 3.3.3

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$^{(a)}$_{I also read that the local solutions can be glued in such a way to build a global solution, but even if that's the case, I've never seen it stated along with the Picard–Lindelöf theorem.}