Where does the second summand come from in the derivative expression? I am fairly certain it has to do with the fact that the boundary is dependent on $t$, but can someone provide either intuitive insight or a formal proof of the fact? I am mostly interested in seeing where the $-c$ comes from since I know its used to prove the finite speed of propogation.
In the integral $E(t)$ the integration domain changes with $t$, so extra care is to be taken when one computes the derivative. Compute $E(t+\mathrm dt)$ where $\mathrm dt$ is an infinitesimal number to the first order in $\mathrm dt$.
Then when you compute $E(t+\mathrm dt)-E(t)$ you need to separate the integration domain into $\mathcal B=\{x:|x-x_0|\leq c(t_0-t-\mathrm dt)\}$ and $\mathcal S=\{x:c(t_0-t-\mathrm dt)\leq |x-x_0|\leq c(t_0-t)\}$: $$E(t+\mathrm dt)-E(t)= \frac12\int_{\cal B} (u_t^2+c^2|Du|^2)\big|_{t+\mathrm dt} -\frac12\int_{\cal B} (u_t^2+c^2|Du|^2)\big|_t -\frac12\int_{\cal S} (u_t^2+c^2|Du|^2)\big|_{t+\mathrm dt}.$$ So it is legitimate to derivate under the integral for the ${\cal B}$ integral, but a second term exists, the integral over $\cal S$. $\cal S$ is a thickened sphere of thickness $c\mathrm dt$, this means that is sufficient to expand the integrand to 0$^{\text{th}}$ order to get the first order $\mathrm dt$ expansion. We have $$E(t+\mathrm dt)-E(t)=\left(\int_{\cal B}u_tu_{tt}+c^2Du Du_t\right)\mathrm dt -\left(\frac12\int_{\cal S}u_t^2+|Du|^2\right)c\mathrm dt $$ which is the formula in the textbook.