Assume $ U\subset V\subset \mathbb{R}^n$ are bounded open subsets with smooth boundary. We define $H^{-s}(\Omega)=(H_0^{s}(\Omega))'$ for $s>0$.
It is straightforward to show that $\left. v\right|_{U}\in H^s(U)$ for all $v\in H^s(V)$ when $s\geq 0$ and that this is continuous.
However, it is not clear to me how this should work for negative orders. I think I would need a continuous extension map $\iota: H_0^s(U)\rightarrow H_0^s(V)$, then I could set $\left. v\right|_{U}=v\circ \iota $ for $v\in H^{-s}(V).$ Intuitively I would define $\iota$ as the extension by $0$, but when looking this up in "Non-Homogeneous Boundary Value Problems and Applications" by Lions and Magenes I saw, that extension by $0$, $H_0^s(\Omega)\rightarrow H^s(\mathbb{R}^n)$ is only continuous for $s \not = \text{integer}+\frac{1}{2}.$
This is probably due to the fact $H_0^{\frac{1}{2}}(\Omega)=H^{\frac{1}{2}}(\Omega)$ for Lipschitz domains, e.g. $1\in H_0^{\frac{1}{2}}(\Omega)$, but indicator functions are not in $H^{\frac{1}{2}}$.
So is there any other obvious way this works?
For your definition of negative order Sobolev spaces, restriction will not always be possible for $s=-1/2,-3/2,\dots$.
This is remedied by slightly altering the definition: For $s\ge 0$ define $H^s_{\bar \Omega}(\Omega)\subset H_0^s(\Omega)$ as the space of functions for which the zero extension lies in $H^s(\mathbb{R^n})$ (here we have $"=" $ if $s\neq 1/2,3/2,\dots$, as you noted correctly). Then define $$ H^{-s}(\Omega):=H_{\bar \Omega}^s(\Omega)^*,\quad s\ge 0.\tag{1} $$ A restriction operator can be constructed by the method you suggest; in particular, we have $$ H^{s}(\Omega)=\{u\in \mathcal{D}'(\Omega):u=U\vert_{\Omega} \text{ for some } U\in H^s(\mathbb{R}^n)\},\quad s\in \mathbb{R}. \tag{2} $$ One can also view (2) as definition (valid for all $s\in \mathbb{R}$), that does not require any interpolation or duality. This viewpoint is taken in Taylor's PDE book; then (1) becomes a theorem (Exercise 18 in Chapter 4.5).
As an illustration, consider $H^{-1/2}(\Omega)$ (with the definition from above) vs. $H^{-1/2}_*(\Omega)=H^{1/2}_0(\Omega)^*$ (your definition). Note that $H^{-1/2}_*(\Omega)\subset H^{-1/2}(\Omega)$ is a proper subset; so for nested opens $U\subset V\subset \mathbb{R}^n$ we can always restrict $f\in H^{-1/2}_*(V)$ to an element $f\vert_U\in H^{-1/2}(U)$, but it is not guaranteed that $f\vert_U$ will lie in the smaller space $H^{-1/2}_*(U)$. This is indeed reasonable: the stronger condition $f\in H_*^{-1/2}(V)$ means that one can make sense of $\langle f, \varphi\rangle$, even for test functions $\varphi\in H^{1/2}_0(V)=H^{1/2}(V)$; this poses a condition on the behaviour of $f$ at the boundary $\partial V$. Of course, there is no reason to assume that $f\vert_{U}$ has a similar boundary behaviour at $\partial U$; after all we are talking about a different boundary now.