Interpretation of the Lagrangian.
I will use $t$ instead of $Z$ and denote $u_Z$ by $\dot u$,
since the equations look more familar in this case.
We define $\mathcal L(u,\nabla u, \dot u, \Phi, \nabla \Phi, \dot \Phi)$ via
$$
\mathcal L = \ i \big(u^* \dot u - u \dot u^* \big) - | \nabla u |^2 + 4\Phi | u |^2 - \nu | \nabla \Phi |^2 + 2q |\Phi|^2.
$$
Hamiltons principle from classical field theory now implies that we look for stationary points of
$$S(u,\Phi) = \int_{\Omega_T} \mathcal L(u,\nabla u, \dot u, \Phi, \nabla \Phi, \dot \Phi) \mathrm{dvol}.$$
($\Omega_T$ denotes the domain of $u$ and $\Phi$, for example $[0,T] \times \Omega$.)
The Euler-Lagrange-Equations are given as
$$ \frac{\partial \mathcal L}{\partial u} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot u} - \sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L}{\partial \mathrm D_i u} = 0$$
and
$$ \frac{\partial \mathcal L}{\partial \Phi} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot \Phi} - \sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L}{\partial \mathrm D_i \Phi} = 0.$$
The principle idea behind these equations is partial integration, basically the same as here: energy minimization.
Here $\mathrm D_i u = \frac{\partial u}{\partial X_i}$ and the derivative $\frac{\partial \mathcal L}{\partial \mathrm D_i u}$ denotes the derivative of $\mathcal L$ with respect to the components of $\nabla u$ in $\mathcal L$.
(To make this mathematical sound, the concept of jets is helpful, but I don't want to go into this. Basically, we consider $u$, $\dot u$ and $\nabla u$ to be independent when computing the derivatives.)
The rest is 'just' a computational task...
The influence of $i (u^* u_Z - u u_Z^*)$?
In my notation, this term is denoted by $i (u^* \dot u - u \dot u^*)$.
Let me split real and imaginary parts, $u = v + iw$, then this term becomes
$$\begin{align}
i \left( (v-iw)(\dot v + i \dot w) - (v+iw)(\dot v - i \dot w)\right) &= 2 i ~ \text{Im}( (v-iw)(\dot v + i\dot w) ) \\
&= -2 ( v \dot w - w \dot u )\\
&= 2 ( w \dot v - v \dot w ). \end{align}$$
This term enters the Euler-Lagrange equations trough
$$
\begin{align}
\frac{\partial \mathcal L}{\partial u} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot u}
&= \frac{\partial \mathcal L}{\partial v} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot v} +
i \frac{\partial \mathcal L}{\partial w} - i\frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot w}\\
&= -2 \dot w - \frac{\mathrm d}{\mathrm d t}(2w) + i (2 \dot v - i \frac{\mathrm d}{\mathrm d t}( -2 w ) )\\
&= 4i \frac{\partial u}{\partial t}.
\end{align}$$
The rest...
The complete computation is a bit to long for me to write down here, but not complicated anymore. Therefore, I just demonstrate one step.
For example a term like $\mathcal L_1 = |\nabla \Phi|^2$ leads to the term
$$-\sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L_1}{\partial \mathrm D_i \Phi}
= -\text{DIV}( 2 \nabla \Phi ) = -2 \Delta \Phi$$
in the Euler-Lagrange equations.
At the end, the first equation is divided by $4$ and the second equation by $2$.
