It is not appropriate to work in the space $H^1_0(U)$ since nonzero boundary conditions are being considered. You will have to assume some regularity of the boundary of $U$.
One version of Green's theorem (see e.g. the appendix in Evans) is that $$- \int_U (\Delta u) v \, dx = \int_U Du \cdot Dv \, dx -\int_{\partial U} \frac{\partial u}{\partial \nu} v \, dS.$$
A weak solution to the problem at hand can be proposed by setting $-\Delta u = f$ in $U$ and $\dfrac{\partial u}{\partial \nu} = -u$ on $\partial U$ so that $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} uv \, dS \quad \forall v \in H^1(U)$$
or a bit more precisely
$$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS \quad \forall v \in H^1(U)$$
where $\tr : H^1(U) \to L^2(\partial U)$ is the trace operator.
An appropriate bilinear form is thus given by $$B[u,v] = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS, \quad u,v \in H^1(U).$$
$B$ is clearly bounded. As far as coercivity goes, it may be helpful to use the Rellich-Kondrachov theorem. I can follow up with a hint if you like.
It remains to show that there is a constant $\alpha > 0$ with the property that $\|u\|_{H^1}^2 \le \alpha B[u,u]$ for all $u \in H^1(U)$. This can be proven by contradition. Otherwise, for every $n \ge \mathbb N$ there would exist $u_n \in H^1(U)$ with the property that $\|u_n\|^2_{H^1} > n B[u_n,u_n]$. For each $n$ define $v_n = \dfrac{u_n}{\|u_n\|_{H^1}}$. Then $v_n \in H^1(U)$, $\|v_n\|_{H^1} = 1$, and $B[v_n,v_n] < \dfrac 1n$ and all $n$.
Here we can invoke Rellich-Kondrachov. Since the family $\{v_n\}$ is bounded in the $H^1$ norm, there is a subsequence $\{v_{n_k}\}$ that converges to a limit $v \in L^2(U)$. However, since $\|Dv_{n_k}\|_{L^2}^2 < \dfrac{1}{n_k}$ it is also true that $Dv_{n_k} \to 0$ in $L^2$. Thus for any $\phi \in C_0^\infty(U)$ you have $$\int_U v D \phi \, dx = \lim_{k \to \infty} \int_U v_{n_k} D \phi \, dx = - \lim_{k \to \infty} \int_U D v_{n_k} \phi \, dx = 0.$$ This means $v \in H^1(U)$ and $D v = 0$, from which you can conclude $v_{n_k} \to v$ in $H^1(U)$. Since $\|v_{n_k}\|_{H^1} = 1$ for all $k$ it follows that $\|v\|_{H^1} = 1$ as well.
Next, since $\|\tr v_{n_k}\|_{L^2(\partial U)}^2 < \dfrac{1}{n_k}$ and the trace operator is bounded there is a constant $C$ for which
$$ \|\tr v\|_{L^2(\partial \Omega)} \le \|\tr v - \tr v_{n_k}\|_{L^2(\partial \Omega)} + \|\tr v_{n_k}\|_{L^2(\partial \Omega)} < \frac{1}{n_k} + C \|v - v_{n_k}\|_{H^1(U)}.$$
Let $k \to \infty$ to find that $\tr v = 0$ in $L^2(\partial U)$.
Can you prove that if $v \in H^1(U)$, $Dv = 0$, and $\tr v = 0$, then $v = 0$? Once you have established that fact you arrive at a contradiction, since $v$ also satisfies $\|v\|_{H^1} = 1$. It follows that $B$ is in fact coercive.
Formula \eqref{2} you give includes the solution to the Dirichlet problem for both equations, i.e. for Laplace's and Poisson's equations.
Precisely, the Dirichlet problem for Poisson's equation reads as
$$
\begin{cases}
\Delta u(x)=f(x) & x\in \Omega\\
u|_{\partial\Omega}=U(x) & x\in \partial\Omega
\end{cases},
$$
while the same problem for Laplace’s equation looks identical except for the fact that in this case $f\equiv0$.
If $f(x)\not\equiv 0$ then $\Delta u(x)\not\equiv 0$ and you have the proper Poisson's equation, thus
$$
\int_\Omega G(x,\xi)\Delta u \,dy\neq 0,
$$
and formula \eqref{2} holds in full. On the other hand, if $f(x)\equiv 0$ then $\Delta u(x)\equiv 0$ thus you have the proper Laplace's equation and
$$
\int_\Omega G(x,\xi)\Delta u \,dy= 0,
$$
thus formula \eqref{2} reduces to formula \eqref{1}.
Best Answer
Evans proves the representation formula in general (see Theorem 12 p. 35). The theorem states that if $u\in C^2(\bar{U})$ solves the boundary value problem and if Green's function exists, then the representation formula holds.
However, at that point in the book the tools for showing the existence of such an $u$ and $G$ are not yet developed. The required tools are introduced in Chapter 6 in Evans book. Basically, they state that if the boundary of your domain $U$ is regular enough, then such a $G$ exists. If additionally $f,g$ also satisfy some requirements (like Hölder continuity) then such a solution $u$ exists. Nonetheless, these tools only give you the general existence of these functions $u,G$. They don't tell you what they actually look like.
Evans constructs the Green function $G$ in the book for two special cases. However, since we don't know at this point that such solutions $u$ exists we can not apply Theorem 12 to show that the representation formula holds. For example at this point in the book we don't know that a $u$ exists such that $\Delta u = 0$ in $\mathbb{R}^n_+$ with $u(x) \rightarrow u(x^0)$ for $x\rightarrow x^0 \in \partial \mathbb{R}^n_+$, hence we can not apply Theorem 12 to get Theorem 14. Thus, we have to actually verify that our constructed $u$ is a solution to our problem.