For the first doubt, the basic idea is that it is an application of the "chain rule" of differentiation. To illustrate: Let $F(t,x)$ be a function, and write by $F'(t,x) = \partial_x F$. Let $x_0(t)$ and $x_1(t)$ be functions. Define
$ E(t) = \int_{x_0(t)}^{x_1(t)} F'(t,s)\mathrm{d}s = F(t,x_1(t)) - F(t,x_0(t))$
by the fundamental theorem of calculus, then you have
$ \frac{d}{dt} E(t) = \left.\partial_tF(t,\cdot)\right]_{x_0(t)}^{x_1(t)} + F'(t, x_1(t))\frac{d}{dt}x_1(t) - F'(t,x_0(t)) \frac{d}{dt}x_0(t) $
For sufficiently nice functions, we can commute the partial derivative with the integral sign and write
$ \partial_t F(t,\cdot) ]_{x_0}^{x_1} = \int_{x_0(t)}^{x_1(t)} \partial_t F'(t,s) \mathrm{d}s $
using the fundamental theorem of calculus again. Hence we have that
The time derivative of an integral whose boundary is time dependent is equal to the integral of the time derivative of the integrand plus boundary terms.
In your specific case I suggest that you compute by changing variables. First note that without loss of generality we can set $x_0, t_0$ in your expression to be both $0$, and take $t < 0$. Then
$ e(\tau) = \int_{B_0(|\tau|)} u_t^2 + |Du|^2 \mathrm{d}x = |\tau|^n \int_{B_0(1)} u_t^2(\tau, |\tau| y) + |D u(\tau,|\tau| y)|^2 \mathrm{d}y $
Now you can take derivative in $\tau$: since the domain of integration is now fixed, there is no contribution from the moving boundary, and you can take the derivative inside the integral (after using the product rule). Then it is a bit of manipulations with chain rules and integration by parts that leads you to the final expression. (Hints:
$ \frac{d}{d\tau} u_t^2(\tau,|\tau|y) = 2 u_t u_{tt} + 2 u_t (y\cdot Du_t)(\tau, |\tau| y) $
and
$ (Du_t)(\tau, |\tau| y) = \frac{1}{|\tau|} D_y( u_t(\tau,|\tau|y)) $
can be useful; the term when the derivative hits $|\tau|^n$ is used to compensate when you integrate by parts and $D$ hits on the $y$ outside.)
For the second doubt, yes, it is just applying the wave equation.