Let $V \subset H \subset V$ be Hilbert triple. We have $u_m$ is infinite differentiable from $[0,T]$ to $V$.
Suppose $u_m \to u$ in $L^2(0,T;V)$ and $u_m' \to u'$ in $L^2(0,T;V^*)$
Suppose that it is true that $$\frac{d}{dt}|u_m(t)|_H^2 = 2\langle u'_m(t), u_m(t) \rangle_{V^*, V}\tag{1}$$ for each $m$.
From the convergences above, it follows that $$|u_m|^2_H \to |u|^2_H$$ and $$\langle u'_m(t), u_m(t) \rangle_{V^*, V} \to \langle u'(t), u(t) \rangle_{V^*, V}$$ both in in $L^1(0,T)$.
These convergences also hold in the distribution sense; therefore we are allowed to pass to the limit in (1) in the distribution sense; in the limit we find precisely $$\frac{d}{dt}|u(t)|_H^2 = 2\langle u'(t), u(t) \rangle_{V^*, V}$$
How does this work exactly? What does he mean by "also hold in distribution sense?" How to pass to the limit in the $\frac{d}{dt}$ term?
Take a function $\varphi \in C^\infty([0,T])$. By (1) you know $$\int_0^T \frac{d}{dt}|u_m|_H^2 \, \varphi \, dt = 2 \, \int_0^T \langle u_m', u_m \rangle_{V^*,V} \, \varphi \, dt.$$ Integrating by parts yields $$-\int_0^T |u_m|_H^2 \, \frac{d}{dt} \varphi dt = 2 \, \int_0^T \langle u_m', u_m \rangle_{V^*,V} \, \varphi \, dt.$$ Now, you can pass to the limit $m \to \infty$ and obtain $$-\int_0^T |u|_H^2 \, \frac{d}{dt} \varphi dt = 2 \, \int_0^T \langle u', u \rangle_{V^*,V} \, \varphi \, dt.$$ This yields $$\int_0^T \frac{d}{dt}|u|_H^2 \, \varphi \, dt = 2 \, \int_0^T \langle u', u \rangle_{V^*,V} \, \varphi \, dt.$$ Since $\varphi \in C^\infty([0,1])$ was arbitrary, this shows $$\frac d{dt} |u|_H^2 = \langle u', u\rangle_{V^*,V}$$ almost everywhere in $(0,T)$.
The phrase "hold in the distributional" sense means, that we have convergence if we see $|u_m|_H^2$ (and its temporal derivative) and $\langle u_m', u_m\rangle_{V^*,V}$ as distributions. This is (more or less) equivalently to what I have done above.