Does there exist a flow for the outward-pointing normal vector field of the unit square?

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For context, maybe I'll first describe the problem I'm actually interested in, which will make it more clear why I'm interested in this particular sub-case as a potential counter-example.

More General Problem: Consider a bounded, connected, simply-connected open set $\mathcal{R} \subseteq \mathbb{R}^3$. The boundary $\partial\mathcal{R}$ is a closed two-dimensional surface embedded in $\mathbb{R}^3$. Suppose that this surface can be written as $$ \partial\mathcal{R} = S_1 \sqcup \partial S_1 \sqcup S_2 = S_1 \sqcup \partial S_2 \sqcup S_2 $$ for some smooth surfaces $S_1$ and $S_2$. In essence, $\mathcal{R}$ is a closed surface, but is only piecewise smooth. A simple example of such a surface is a circular disk, with a half-sphere covering it on top. Choose outward-pointing unit orientation fields $\mathbf{N}_1$ and $\mathbf{N}_2$ for the smooth surfaces $S_1$ and $S_2$, respectively.

If we consider just one of these vector fields individually, say $\mathbf{N}_1$ for now, then one can smoothly extend this vector field into a neighbourhood of $S_1$ (highly non-unique, but I don't need it to be). There then exists a flow $\varphi_s$ that satisfies $\frac{d\varphi_s(p)}{ds}\big|_{s=0} = \mathbf{N}_1(p)$.

My question is, can we find a flow defined in a neighbourhood of $\partial\mathcal{R}$ that generates both $\mathbf{N}_1$ and $\mathbf{N}_2$ simultaneously?

Special Case: Consider the unit square in $\mathbb{R}^2$ with a unit outward pointing vector field everywhere except at the corners. Does there exist a flow for this vector field?

Since the corners are not removable discontinuities, it doesn't seem to me like this is possible? The hand-waving physicist in me wants to remove an $\epsilon$-ball from the corner, then send $\epsilon \to 0$ to make this happen though.