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.
It suffices to use the space $V=H^1$ along with the trace theorem
Indeed it follows from trace theorem that, the injection $$H^1(\Omega)\to H^{1/2}(\Gamma)\to L^{2}(\Gamma)$$
are continuous therefore, there is a constant k such that for all $u\in H^1(\Omega)$
$$\| u\|_{ L^{2}(\Gamma)}\le k\|u\|_{H^1(\Omega)}$$
from this, you easily get the continuity of the bilinear form $B$ on $H^1(\Omega)\times H^1(\Omega)$ whereas the coercivity is straightforward since, $c(x)>c_0$. By the same token you get the continuity on $H^1(\Omega)$ of the linear form
$$v\mapsto \int_\Omega fv+\int_{\Gamma }gv.$$
therefore the existence and uniqueness for the weak solution follow from Lax-Milgramm
Best Answer
This should follow from local boundary $H^2(\Omega)$ regularity, applied separately near $\Gamma_0$ and $\Gamma_1$. Since the two boundary conditions are prescribed on separated regions (which is what I assume you mean by an annular domain), you can localise to one and ignore the other.
In what follows, note that a weak solution $u$ satisfies the weak formulation $$ \int_{\Omega} \nabla u \cdot \nabla \varphi + u \varphi \,\mathrm{d}x = \int_{\Omega} f \varphi \,\mathrm{d}x + \int_{\Gamma_1} (g-u)\varphi \,\mathrm{d}x$$ for all $\varphi \in H^1_{\Gamma_0}(\Omega)$, that is all $\varphi \in H^1(\Omega)$ vanishing on $\Gamma_0$.
For regularity near $\Gamma_0$, we can take test functions $\varphi \in H^1_0(\Omega)$ and observe that the $\Gamma_1$ term vanishes in the weak formulation. Then $H^2$ regularity follows in the standard way by:
Locally flattening the boundary near some $x_0 \in \Gamma_0$,
Taking tangential difference quotients to show all the tangential derivatives $\partial_iu$ locally lie in $H^1$ for $1 \leq i \leq n-1$. Here it is convenient that the Dirichlet boundary is zero, which ensures the difference quotients remain valid test functions.
Using the equation deduce that the final derivative $\partial_nu$ also locally lies $H^1$.
This is standard, and can be found in any graduate level PDE text covering elliptic equations. See for instance Chapter 6, Section 3.2 of
Evans, Lawrence C., Partial differential equations, Graduate Studies in Mathematics. 19. Providence, RI: American Mathematical Society (AMS). xvii, 662 p. (1998). ZBL0902.35002.
For regularity near $\Gamma_1$, the strategy is similar, involving flattening the boundary and using difference quotients. Here $\varphi$ built from difference quotients will automatically be a valid test function (assuming your cutoff vanishes near $\Gamma_0$), but you need to estimate an additional term coming from the $\Gamma_1$ integral. I will leave this an exercise, but since you are only taking difference quotients in the tangential directions it's just a matter of using the trace theorem to estimate $u$.