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.
I don't know if your approach could be made to work, but in my opinion you're going the wrong way. First of all, since $c=c(x,t)$, you can't just ignore the factor $e^{ct}$ when taking the laplacian of $v$. Moreover, the convolution integral will, in general, not be a solution of your problem: If $g>0$ then the convolution integral defines a strictly positive function on $\mathbb{R}^n$; in particular it can't satisfy the boundary conditions.
What you want to try is the maximum principle: Set $v=e^{\gamma t}u$, then $v_t-\Delta u =(\gamma-c)v$. Since $\gamma-c<0$ we can apply the maximum principle to get, for $(x,t)\in U_T$,
$$
-\max_{U}g^-=-\max_{\Gamma_T}v^-\leq \min_{U_T} v \leq \sup_{U_T} v =\sup_{\Gamma_T} v^+ = \sup_U g^+.
$$
This implies the desired inequality with $C\leq \sup_U |g|$.
Best Answer
Here is a solution more related to Evan's approach in the mentioned textbook. Let $u = u(x,t)$ be a solution of the PDE. Then using integration by parts and that $u_t = \Delta u$ we get
$$\begin{align}\frac{d}{dt} \left(\frac{1}{2} \|u\|_{L^2(U)}^2\right) &= \int_U u_tu\ dx = \int_U u \Delta u\ dx\\&= -\int_U |Du|^2 dx \overset{(\ast)}\leq - \lambda_1 \|u\|^2_{L^2(U)}\end{align}$$
where $(\ast)$ comes from Rayleigh's Formula
$$\lambda_1 = \underset{\substack {u \in H_0^1 (U)\\ u\neq 0}}\min \frac{B[u,u]}{\|u\|^2_{L^2(U)}} = \underset{\substack{u \in H_0^1 (U)\\ u\neq 0}} \min \frac{\int_U |Du|^2 dx}{\|u\|^2_{L^2(U)}} $$
Now let $\eta (s) = \|u(\cdot,s)\|^2_{L^2(U)}$. Then
$$\frac{d}{ds} \left(\eta(s) e^{2\lambda_1 s}\right) = e^{2\lambda_1 s} (\eta'(s) +2\lambda_1 \eta(s)) \leq 0$$ Integrating from $0$ to $t$ w.r.t. $s$ we obtain
$$\eta(t)e^{2\lambda_1 t} \leq \eta (0)$$
Since $\eta (0) = \|u (\cdot, 0)\|^2_{L^2(U)} = \|g\|^2_{L^2(U)}$ the result follows.