= Solution
Let $v_i=\dot x_i$ and use a fixed arbitrary material subvolume $B_0$ in the <reference configuration>, with outward unit normal $N_I$. The reference mass density $\rho_0$ is time independent. The global <momentum conservation> law is
$$
\frac d{dt}\int_{B_0}\rho_0v_i\,dX
=\int_{\partial B_0}P_{Ii}N_I\,dS+\int_{B_0}\rho_0g_i\,dX.
$$
The <divergence theorem>, differentiation under the integral, and arbitrariness of $B_0$ give
$$
\boxed{\rho_0\dot v_i=P_{Ii,I}+\rho_0g_i.}
$$
The global <conservation of energy> law includes both <internal energy> and <kinetic energy>:
$$
\frac d{dt}\int_{B_0}\rho_0\left(u+\frac12v_iv_i\right)dX
=\int_{\partial B_0}(P_{Ii}v_i-q_I^0)N_I\,dS
+\int_{B_0}\rho_0(g_iv_i+r)\,dX.
$$
Here $q^0$ is outgoing <heat flux> per reference area. Localizing gives
$$
\rho_0\dot u+\rho_0v_i\dot v_i
=P_{Ii,I}v_i+P_{Ii}v_{i,I}-q_{I,I}^0
+\rho_0g_iv_i+\rho_0r.
$$
Subtract the scalar product of the local <momentum conservation> law with $v$. Since $v_{i,I}=\dot F_{iI}$, this proves the <Lagrangian internal energy balance>
$$
\boxed{\rho_0\dot u=P_{Ii}\dot F_{iI}+\rho_0r-q_{I,I}^0.}
$$
For positive <temperature> $\theta$, the global <entropy> inequality is
$$
\frac d{dt}\int_{B_0}\rho_0\eta\,dX
\geq\int_{B_0}\frac{\rho_0r}{\theta}\,dX
-\int_{\partial B_0}\frac{q_I^0N_I}{\theta}\,dS.
$$
Its local excess is a nonnegative production density $\gamma$. Expanding the divergence of $q^0/\theta$ proves the <Lagrangian entropy inequality> in the requested equality form:
$$
\boxed{\rho_0\dot\eta
=\frac{\rho_0r-q_{I,I}^0}{\theta}
+\frac{\theta_{,I}q_I^0}{\theta^2}+\gamma,
\qquad\gamma\geq0.}
$$
No constitutive assumption on the <heat flux> was needed for these balance identities.
For the <reference-configuration jump balances>, make the orientation convention explicit. Write $[a]=a^+-a^-$ and let $n^0$ point from the minus side to the plus side. Let $c$ be the actual interface speed in direction $n^0$. A signed level function $\phi$ with the plus side $\phi>0$ obeys $\phi_t=-c|\nabla_X\phi|$ on the interface. For $a=a^-+[a]H(\phi)$, the singular time derivative is $-c[a]\delta_S$; the singular reference divergence of $b$ is $n_I^0[b_I]\delta_S$. Thus a local conservation law $a_t=\operatorname{Div}b+$ bounded sources has the <Rankine-Hugoniot condition>
$$
-c[a]=n_I^0[b_I].
$$
This is also the shrinking moving-pillbox calculation of the global law: finite volume supplies have no surviving surface term.
To match the displayed positive-sign momentum jump in the paper, use its speed parameter as $V=-c$, so the interface travels in direction $-Vn^0$. Assuming continuous $\rho_0$ and no surface force or energy supply, the jump laws are then
$$
\boxed{\rho_0V[v_i]=n_I^0[P_{Ii}],}
$$
$$
\boxed{\rho_0V\left[u+\frac12v_iv_i\right]
=n_I^0[P_{Ii}v_i-q_I^0],}
$$
and, from $\partial_t(\rho_0\eta)+\operatorname{Div}(q^0/\theta)\geq\rho_0r/\theta$,
$$
\boxed{\rho_0V[\eta]+n_I^0[q_I^0/\theta]\geq0.}
$$
The last inequality is nonnegative entropy production concentrated on the interface. Temperature and flux in the entropy term are the traces on their respective sides; one cannot take a common temperature outside the jump unless it is continuous.
If $V$ is instead defined as the interface speed in direction $n^0$, all three $V$ terms above acquire a minus sign. In particular the printed positive-sign momentum formula requires the opposite speed convention; changing just the order of the jump does not remove this convention issue. If reference density is discontinuous, replace $\rho_0[a]$ by the jump $[\rho_0a]$ of the complete conserved density.
An equivalent internal-energy jump follows by subtracting the kinetic-energy jump using $[|v|^2/2]=\{v_i\}[v_i]$, where $\{a\}=(a^++a^-)/2$:
$$
\rho_0V[u]=n_I^0\{P_{Ii}\}[v_i]-n_I^0[q_I^0].
$$
The averaged traction appears because $[P_{Ii}v_i]=\{P_{Ii}\}[v_i]+[P_{Ii}]\{v_i\}$; using a one-sided traction would generally be incorrect.
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