From Evans Partial Differential Equation:
(Integration by parts formula). Let $u,v \in C^1(\bar U)$. Then $$\int_U u_{x_i} v dx = -\int_U v_{x_i}u dx + \int_{\partial U}uv\nu^idS \ \ \ (i= 1,... , n).$$
My question is: Since $u,v \in C^1(\bar U)$, it means $u,v \in C^1(U)$ and $u, v, Du, Dv$ are uniformly continuous on bounded subsets of $U$, so the domain of $u,v$ is $U$ only, the domain does not include $\partial U$ (boundary of $U$). Then how to evaluate $\int_{\partial U}uv\nu^idS$?
You are correct, $u,v \in C^1(\bar{U})$ does mean $u,v,Du,Dv$ are uniformly continuous on $U$ (here, Evans assumes $U$ is bounded, so the "bounded" qualifier is unneeded) and $\partial U$ is not explicitly mentioned. It is a fact (from basic analysis) that every uniformly continuous function $f \in C(\bar{U})$ has a unique extension to a continuous function $f:\bar{U} \to \mathbb{R}$. If you are not familiar with this "fact", I can add details. This extension is what is used to define the integral over the boundary.