How did the second term become zero?
Recall that the cross product of vectors $\vec{a}$ and $\vec{b}$ is a vector representing the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$ in the direction perpendicular on the parallelogram.
$$
\begin{align}
\vec{\omega}(t) \times \mathbf{J}\vec\omega(t) &=
\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}
\times
\begin{pmatrix} J_x & 0 & 0 \\ 0 & J_y & 0 \\ 0 & 0 & J_z \end{pmatrix}
\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix} \\
&=
\underbrace{\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}}_{\vec{a}}
\times
\underbrace{\begin{pmatrix} 0 \\ J_y\omega_y \\ 0 \end{pmatrix}}_{\vec{b}}
\\
&= \vec{0}
\end{align}
$$
Note that within this case the vectors $\vec{a}$ and $\vec{b}$ are parallel, hence the area of the parallelogram they span is $0$ and therefore the cross product is also zero.
How did they calculate the external moment vector?
There is only one external force present in this system, the bearing force
$$\vec{F} = \begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}.$$
The moment this forces causes on the center of gravity is
$$\begin{align}
\vec{M}_C
&= \vec{r}_C \times \vec{F}\\
&= \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} \times
\begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}\\
&=
\begin{pmatrix} A_z r_y - 0 r_z \\ A_x r_z - A_z r_x \\ 0 r_x - A_x r_y \end{pmatrix}
\end{align}
$$
Assuming that $r_y = 0$ result in
$$ \vec{M}_C =
\begin{pmatrix} 0 \\ A_x r_z - A_z r_x \\ 0 \end{pmatrix}
$$
Note that the rotation $\varphi_y$ is clockwise and not counter clockwise as it would have been by using a right handed coordinate system.
Therefore the external moment should be defined in the opposite direction
$$ \vec{\mathbf{I}} = -\vec{M}_C =
\begin{pmatrix} 0 \\ -A_x r_z + A_z r_x \\ 0 \end{pmatrix}.
$$