Multiply $(40)$ by $u_t$ and integrate: $$\int_{\mathbb{R}^N}u_t(u_t)_t-\int_{\mathbb{R}^N}u_t\Delta u_t=\int_{\mathbb{R}^N}f_tu_t.\tag{1}$$
Use integration by parts in $(1)$ combined with $(u_t^2)_t/2=u_t(u_t)_t$ to conclude that $$\frac{1}{2}\int_{\mathbb{R}^N} (u_t^2)_t+\int_{\mathbb{R}^N} |Du_t|^2=\int_{\mathbb{R}^N} f_tu_t. \tag{2}$$
Now integrate $(2)$ from $0$ to $T$ to get
\begin{eqnarray}
\int_0^T\int_{\mathbb{R}^N} |Du_t|^2 &=& \int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}u_t^2(\cdot, 0)-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right) \nonumber \\
&=& \int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}(f(\cdot, 0)+\Delta g)^2-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right). \tag{3}\end{eqnarray}
From $(3)$ $$\sup_{s\in[0,T]}\int_{\mathbb{R}^N}|u_t(\cdot,s)|^2+\int_0^T\int_{\mathbb{R}^N} |Du_t|^2=\sup_{s\in[0,T]}\int_{\mathbb{R}^N}|u_t(\cdot,s)|^2+\int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}(f(\cdot, 0)+\Delta g)^2-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right).\tag{T}$$
From here you can proceed as follows. We combine $(2)$ with inequality $2ab\le a^2+b^2$ to obtain $$\int_{\mathbb{R}^N} (u_t^2)_t\le \int_{\mathbb{R}^N}(f_t^2+u_t^2). \tag{4}$$
Let $$\eta(s)=\int_{\mathbb{R}^N} u_t^2(x,s)dx,\ \ \phi(s)=\int_{\mathbb{R}^N}f_t^2(x,s),\ s\in [0,T],$$
and note that from $(4)$ $$\eta'(s)\le \eta(s)+\phi(s).$$
Now apply Gronwall inequality and use $(T)$ to conclude.
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.
Best Answer
I am willing to share some idea. Since $f$ is bounded, we can let $M=:\sup |f|$. Define $$ \begin{cases} -\Delta\bar{u}=M \\ u|_{\partial_{U}}=0\\ \end{cases},\begin{cases} -\Delta\underline{u}=-M \\ u|_{\partial_{U}}=0\\ \end{cases} $$ Using maximum principle, we can directly check $\underline{u}\le\bar{u}$. Then we can obtain the existence of solution. However, we also need to prove the positivity. We only need to prove $\lVert u^{-}\rVert_{L^2(U)}=0$. we know that $$ \int_U Du Dvdx=\int_{U} f(u)vdx ,\quad\forall v\in H_0^1(U) $$ If we let $v=u^{-}$, we obtain $$ -\int_U |Du^{-}|^2dx=\int_{U} f(u)u^{-}dx $$ In addition, $$ \int_U |Du^{-}|^2dx\ge \lambda_1 \int_U |u^{-}|^2dx $$ In the end, we arrive at $$ \int_U (f(u)+\lambda_1 u^{-})u^{-}dx\le 0 $$ I don't know how to continue. However, if Lip$(f)<\lambda_1$, we can get the conclusion.