In my mind, the rotating spin angular momentum can never have a
component on AB. It will always stay perpendicular to AB and will not
contribute in the change in total angular momentum along AB.
I think I am missing something here. All I know is that if TORQUE
ALONG A DIRECTION IS ZERO, ANGULAR MOMENTUM WILL NOT CHANGE ALONG THAT
DIRECTION WHAT I DON'T KNOW IS THAT WHAT WILL HAPPEN IF THE DIRECTION
ITSELF IS MOVING.
You have the right doubt, and when in doubt, best to revert to the fundamentals to seek resolution. Let us recall the basic principles:
Rate of change of angular momentum in an inertial frame is equal to the torque of external forces (assuming torque of internal forces is zero).
A vector can be changed by changing the magnitude or the direction.
Keeping the above two principles in mind, we first choose the inertial frame to be the lab frame (in which the entire assembly is rotating about the vertical with angular speed $\Omega$). Let us station the origin of the lab frame at the center of the gyrocompass which is clearly stationary in the lab frame.
Next, although the direction $AB$ is changing in space, imagine a fixed moment in time $\ t = t_{o}$. At $\ t=t_o$, $AB \ $ will point in a fixed direction in space. The torque equation (principle 1 above) then tells us that the torque in direction $AB$ at $\ t = t_o$ is equal to the instantaneous rate of change of angular momentum in the direction of $AB$. Mathematically, this means calculating $\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$ at $t=t_o$ and taking its projection along $AB$.
As already explained by Kleppner-Kolenkow, the component of torque on the system along axis $AB$ in the lab frame about our chosen origin is zero (they are assuming that the center of mass of the gyroscope is at its geometrical center and no friction on the axle $AB$). So, the only task at hand is to calculate $\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$ at $t=t_o$ along $AB$.
Now, to do the calculation for $\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$, note that the "spin angular momentum" has a vertical as well as a horizontal component. But the horizontal component is precessing about the vertical with angular velocity $\Omega$ (because the entire assembly is rotating about the vertical)! This implies that the "direction" of the horizontal component of "spin angular momentum" in the lab frame is constantly changing. By principle 2 (stated above), this precession leads to a contribution in the expression for $\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$.
I'd now recommend (re-)reading the previous section in the same chapter of this book (probably goes by the name "gyroscope precession"; also check out this for visualization). The essence of that section is that in case of a purely precessional motion $-$ imagine a vector, $\vec{V}$, of fixed length, spinning about a fixed axis with instantaneous angular velocity $\vec{\omega} \ -$ we have
$$ \frac{\mathrm{d}\vec{V}}{\mathrm{d}t} = \vec{\omega} \times \vec{V}$$
In this particular case, $\vec{\omega} = \Omega \ \hat{k}$, and $\vec{V}$ is the horizontal component of the "spin angular momentum" (because remember the entire assembly is spinning about the vertical and so the horizontal component of the "spin angular momentum" is precessing too). The only minor caveat here is that $\vec{V}$ might change in magnitude $-$ however, this contributes nothing in the direction $AB$ because ($\vec{V}$ is directed perpendicular to $AB$). Clearly then, the precessional contribution in the direction $AB \ $ is given by $\Omega L_s \sin \theta $, and happens to be the only other contribution to $\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$ along $AB$ apart from the usual $I_{\perp}\ddot{\theta} \ - $ and this is exactly what Kleppner-Kolenkow are claiming.
Thus, we have,
$$ \boxed{ \bigg(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\bigg)_{t=t_o}\cdot\vec{e}_{AB} = I_{\perp}\ddot{\theta} + \Omega L_s \sin \theta = 0 }$$
where $\vec{e}_{AB}$ is a unit vector in the direction $AB$.
While this heuristically proves the torque equation, I'd still suggest using Euler's equations or explicitly writing out the components of $\vec{L}$ in the lab frame and taking time derivatives in order to not miss other contributions in more complex setups.
Besides this, as explained by others, friction damps this (pendulum like) oscillatory motion in $\theta$, eventually aligning the axis of the gyroscope with the axis about which the platform is spinning ($\theta = 0$).
Note: This problem just illustrates the principle of a gyrocompass $-$ for an actual gyrocompass device the spinning platform is the earth.
Hope this helps.