The work-KE theorem is a mathematical consequence of finding the work due to the net force on a particle. Using the definition of work, as a line integral, we can calculate:
$$
\begin{align}
W_{F_\text{net}} &= \int_{t_i}^{t_f} \vec{F}_\text{net} \cdot \frac{d \vec{r}}{d t} \, dt\\
&= \int_{t_i}^{t_f} m \frac{d \vec{v}}{d t} \cdot \vec{v} \, dt\\
&= \int_{t_i}^{t_f} m \frac{1}{2} \frac{d }{d t}\left( \vec{v} \cdot \vec{v}\right) \, dt\\
& = \frac{m}{2} \left[ \vec{v} \cdot \vec{v} \right]^{t_f}_{t_i}\\
&= \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2
\end{align}
$$
As Kleppner and Kolenkow point out, we can apply this exact same mathematical analysis to Newton's Second Law for a composite object of many particles:
$$
\vec{F}_\text{net, ext} = M \frac{d \vec{V}}{dt}\\
$$
where $\vec{F}_\text{net, ext}$ is the sum of the external forces on
a collection of $N$ particles; $M$ is the total mass, $\sum_n m_n$; and the center-of-mass velocity of the system of particles is defined as:
$$
\vec{V} = \frac{d\vec{R}}{dt} \quad \text{with} \quad \vec{R} = \frac{m_1 \vec{r}_1 + \cdots + m_N \vec{r}_N}{M}
$$
See my recent answer here for how to derive this equation. The same calculation for "net work" shown above can be accomplished using this equation, but in terms of the center-of-mass velocity. As a shorthand, let's call that the "work on the center of mass":
$$
W_\text{com} := W_{F_\text{net, ext}} = \frac{1}{2} M V_f^2 - \frac{1}{2} M V_i^2
$$
So, in the first part of their example, they are simply finding $W_\text{com}$ for the composite object (a wheel) rolling down the incline. The net external force on the object is the sum of friction, normal, and gravity:
$$
\vec{F}_\text{net, ext} = \vec{F}_{fs} + \vec{F}_N + \vec{F}_g
$$
and the work due to any one of these forces can be found by the simpler expression for work, $\vec{F} \cdot \Delta \vec{R}$, because the force is constant over the path. For example, for friction:
$$
W_{F_{fs}} = \int_{t_i}^{t_f} \vec{F}_{fs} \cdot \frac{d\vec{R}}{dt} \, dt = \vec{F}_{fs} \cdot \int_{t_i}^{t_f} \frac{d\vec{R}}{dt} \, dt = \vec{F}_{fs} \cdot \Delta \vec{R} = - F_{fs} \, \ell
$$
With work done by normal zero, we find the total net work (on the center-of-mass of the composite object) to be:
$$
W_\text{com} = W_{F_\text{net, ext}} = \left( F_g \sin \beta - F_{fs} \right) \ell
$$
But, because this work was defined to be that on the center-of-mass, we have neglected kinetic energy of the rotation of this composite object, and the work done to achieve that "internal" energy of the object.
To find this "work associated to the rotational kinetic energy", we can write down "Newton's Second Law for rotation" (of a rigid body):
$$
\tau_\text{net, $z$} = I \alpha_z = I \frac{d \omega_z}{d t}
$$
And then we can manipulate this equation to get something in the form of a Work-KE theorem:
$$
\begin{align}
\tau_\text{net, $z$} &= I \frac{d \omega_z}{d t}\\
\tau_\text{net, $z$} \, \omega_z &= I \frac{d \omega_z}{d t} \, \omega_z \\
\tau_\text{net, $z$} \, \omega_z &= I \frac{1}{2} \frac{d}{d t}\left(\omega_z^2 \right) \\
\int_{t_i}^{t_f} \tau_\text{net, $z$} \, \omega_z \, dt &= \int_{t_i}^{t_f} I \frac{1}{2} \frac{d}{d t}\left(\omega_z^2 \right) \, dt \\
\int_{t_i}^{t_f} \tau_\text{net, $z$} \, \frac{d\theta_z}{dt} \, dt &= I \frac{1}{2} \left[\omega_z^2 \right]_{t_i}^{t_f} \\
\int_{\theta_i}^{\theta_f} \tau_\text{net, $z$} \, d\theta_z &= \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2
\end{align}
$$
We can then define the left-hand side to be the "work associated to the rotational kinetic energy" and get its relationship to rotational kinetic energy:
$$
W_\text{rot} = \int_{\theta_i}^{\theta_f} \tau_\text{net, $z$} \, d\theta_z = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2
$$
So, what is the connection between these things? It seems, and Kleppner and Kolenkow claim without proof, that the net work on an object should be equal to the "work on the center of mass" added to the "work associated to rotational kinetic energy":
$$
W_\text{net, tot} = W_\text{com} + W_\text{rot}
$$
Is that actually true?
Well, the laws of nature are Newton's Second Law (for a particle) and Newton's Third Law (for interaction between particles). So, if it is true, we should be able to show it from the total net work on particles, i.e., the sum of the net works for all particles in the object. Let's try to calculate that directly:
$$
W_\text{net, tot} = \sum_{i=1}^N W_{F_\text{net, i}} = \sum_i^N \int_{t_i}^{t_f} \vec{F}_\text{net, i} \cdot \vec{v}_i \, dt
$$
For an object that rotates rigidly about its center of mass, we can write the velocity of any particle as $\vec{v}_i = \vec{V} + \vec{v}_\text{rot, i}$, where $\vec{V}$ is again the center-of-mass velocity, and $\vec{v}_\text{rot, i}$ is the velocity of the particle relative to the center of mass (tangent to the circular path of that particle about the rotational axis). Then:
$$
\begin{align}
\sum_{i=1}^N W_{F_\text{net, i}} &= \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, i} \cdot \vec{v}_i \, dt \\
& = \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \left(\vec{V} + \vec{v}_\text{rot, $i$}\right)\, dt \\
& = \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \vec{V}\, dt + \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \vec{v}_\text{rot, $i$}\, dt \\
& = \int_{t_i}^{t_f} \left( \sum_{i=1}^N \vec{F}_\text{net, $i$} \right) \cdot \vec{V}\, dt + \sum_{i=1}^N \int_{t_i}^{t_f} m_i \, \vec{a}_i \cdot \vec{v}_\text{rot, $i$} \, dt
\end{align}
$$
Now, in the first term, Newton's Third Law will lead to a cancellation of all of the (internal) forces of interaction between pairs of particles; the sum of the $\vec{F}_\text{net, i}$'s is thus the sum of the external forces on the object, $\vec{F}_\text{net, ext}$. Therefore:
$$
\int_{t_i}^{t_f} \left( \sum_{i=1}^N \vec{F}_\text{net, $i$} \right) \cdot \vec{V}\, dt = \int_{t_i}^{t_f} \vec{F}_\text{net, ext} \cdot \vec{V} \, dt = W_\text{com}
$$
So the first term gives us exactly what we needed.
To show that the second term is $W_\text{rot}$, recognize that:
$$
\left|\vec{v}_\text{rot, $i$} \right| = \omega R_i \quad \text{and} \quad \vec{a}_i \cdot \vec{v}_\text{rot, $i$} = a_\text{tan, $i$} \,v_\text{rot, $i$}
$$
where $R_i$ is the distance of the $i$th particle from the rotational axis and $a_\text{tan, $i$} = R_i \alpha$ is the tangential component of the acceleration of the $i$th particle. Therefore:
$$
\begin{align}
\sum_{i=1}^N \int_{t_i}^{t_f} m_i \,\vec{a}_i \cdot \vec{v}_\text{rot, $i$} \, dt &= \sum_{i=1}^N \int_{t_i}^{t_f} m_i \left(R_i \alpha\right) \left(\omega R_i \right) \, dt\\
& = \sum_{i=1}^N \int_{t_i}^{t_f} m_i R_i^2 \frac{d \omega}{dt} \omega \, dt\\
& = \int_{t_i}^{t_f} \left( \sum_{i=1}^N m_i \, R_i^2 \right) \frac{1}{2} \frac{d}{dt} \left(\omega^2\right) \, dt\\
& = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 \\
& = W_\text{rot}
\end{align}
$$
Indeed, for an object that rigidly rotates about its center of mass:
$$
W_\text{net, tot} = W_\text{com} + W_\text{rot}
$$
as they were defined by Kleppner and Kolenkow.
