The Euler equations all refer to the body-fixed reference frame. The general equation in this case reads
$$\mathrm{D}_t \vec{J}'=\dot{\vec{J}}'+\vec{\omega}' \times \vec{J}'=\vec{M}',$$
where I've put a prime on the vector symbols to indicate that these are components wrt. the non-inertial body-fixed system. ##\vec{M}'## are the corresponding components of the torque. If the only force acting on the body is the homogeneous gravitational field of the Earth i.e., ##\vec{F}=m \vec{g}## for a point particle, then ##\vec{M}## is given as ##m \vec{s} \times \vec{g}##, where ##\vec{s}## is the vector from the body-fixed point of rotation to the center of mass. If the body is freely falling or if the body-fixed point of rotation is chosen to be the center of mass, then ##\vec{M}'=0## and this gives the equation of motion (9.46) in the book, since
$$\vec{J}'=\hat{\Theta}' \vec{\omega}' \; \Rightarrow \; \hat{\Theta}' \dot{\vec{\omega}}'+\vec{\omega}' \times \hat{\Theta}' \vec{\omega}'.$$
Since ##\hat{\Theta}'## are the components of the tensor of inertia with respect to the body-fixed reference basis around the body-fixed point of rotation you can always choose this basis such to make ##\hat{\Theta}'=\mathrm{diag}(I_1,I_2,I_3)##, and then writing out the equation in components, you get (9.46), which however is for the special case of a symmetric top, where (at least) two of the principle moments of inertia are equal, i.e., as stated in the book ##I_1=I_2=I##.
The free symmetric top is not too difficult to solve. For a full understanding, you should introduce Euler angles between the body- and space-fixed reference bases, choosing the space-fixed (inertial) basis such that the 3-axis points into the direction of the conserved angular momentum.
