I am assuming you want $V$ to actually be the image of $U$. In this case, there is no such map satisfying your second condition.
If $m = n$, this follows from invariance of domain, since the image of $U$ will necessarily be open.
If $m < n$, there is no continuous injective map from $\mathbb{R}^n$ to $\mathbb{R}^m$ (let alone to a closed subset). You can find some more elementary arguments here, or you can again apply invariance of domain. In particular, if $f : U \to V$ is continuous and injective, and $\iota : \mathbb{R}^m \to \mathbb{R}^n$ is an inclusion map, then $\iota \circ f : U \to \mathbb{R}^n$ is an open map, but the image of $U$ is not open in $\mathbb{R}^n$.
For $m > n$, similar logic: we cannot have a continuous injective map from $U$ onto a set with non-empty interior in $\mathbb{R}^m$ (since $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic).
Since we didn't even use the fact that an interior point maps to the boundary of $V$, I suspect there is an easier argument. (Maybe take an interior point $u$ which maps to the boundary, restrict to a small compact neighborhood of $u$, and use the fact that the map on the compact neighborhood is a homeomorphism and must preserve the boundary).
Also, I guess it's possible that you never intended for $V$ to actually be the image of $U$. In this case, we still cannot find such a function when $m \leq n$, based on the same arguments as above, but we can for $m > n$. For example, take $V$ to be the unit disk in $\mathbb{R}^2$ and consider the map $(-1/2,1/2) \to \mathbb{R}^2 : x \mapsto (x,1-4x^2)$, or something similar. This maps $0$ to the boundary of $V$.
DISCLAIMER: I'm not terribly familiar with topology so I don't know whether I've been rigorous enough with this answer. I think it brings the idea across if nothing else.
The set $\mathcal{N}$ can be characterised as the set:
$$
\mathcal{N}=\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}:a^2+b^2+c^2+d^2=1\quad\land\quad ad-bc=0\right\}
$$
If we rewrite these conditions a bit, we see that they are:
$$
(a+d)^2+(a-d)^2+(b+c)^2+(b-c)^2=2\\
(a+d)^2-(a-d)^2-(b+c)^2+(b-c)^2=0
$$
Introducing
$$
x=a+d\quad y=a-d\quad z=b+c\quad w=b-c
$$
we get:
$$
x^2+y^2+z^2+w^2=2\\
x^2-y^2-z^2+w^2=0
$$
But this system is equivalent to
$$
x^2+w^2=1\\
y^2+z^2=1
$$
These solutions can be parametrised as:
$$
x=\cos t\quad w=\sin t\\
y=\cos s\quad z=\sin s
$$
The map $(s,t)\rightarrow (x(s,t),y(s,t),z(s,t),w(s,t))$ is locally a homeomorphism as it essentially just represents a parametrisation of two independent circles with radius $1$ by the angles $s,t$. From here, we recover $a,b,c,d$ as:
$$
a=\frac{\cos t+\cos s}{2}\quad b=\frac{\sin t+\sin s}{2}\quad c=\frac{-\sin t+\sin s}{2}\quad d=\frac{\cos t-\cos s}{2}\quad
$$
But the map $(x,y,z,w)\rightarrow (a,b,c,d)$ is just a linear transformation and as such also a homeomorphism. The composition $(s,t)\rightarrow(x,y,z,w)\rightarrow(a,b,c,d)\rightarrow\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is precisely the map $F$, so $F$ is indeed a local homeomorphism from $\mathbb{R}^2$ to $\mathcal{N}$.
Best Answer
Here's a sketch of another argument along the line of Daniel Fischer's.
By the connectedness of $\mathbb{R}^n$, it suffices to show $U$ is closed. So suppose $x_n \in U$ and $x_n \to x$. Then $\{x_n\}$ is Cauchy, and it follows from the uniform continuity that $\{f(x_n)\}$ is Cauchy as well. $\mathbb{R}^n$ is complete, so $f(x_n)$ converges to some $y \in \mathbb{R}^n$. By the continuity of $f^{-1}$, we have $x_n = f^{-1}(f(x_n)) \to f^{-1}(y)$ so $x = f^{-1}(y)$. In particular, $x \in U$.
This would work just as well if we replaced $\mathbb{R}^n$ by another connected complete metric space.