You are asking about the rotational kinematics of two connected bodies.
Consider two bodies in motion, each having a local coordinate frame attached to it, as well as the ground inertial frame.
The relative rotational motion between the two bodies is defined based on the pin axis $\hat{z}_{\rm pin}$ as seen above, and the joint speed $\dot{\theta}$.
When the relative rotational velocity is expressed in the parent coordinate frame it is $$\vec{\omega}_{\rm rel}^{\{1\}} = \hat{z}_{\rm pin}^{\{1\}} \dot{\theta}$$ and when it is expressed in the inertial frame, it is similarly $$\vec{\omega}_{\rm rel}^{\{0\}} = \hat{z}_{\rm pin}^{\{0\}} \dot{\theta} \tag{1}$$
Just so to make the difference clear, the local pin axis might be an elementary vector like $\hat{z}_{\rm pin}^{\{1\}} = \pmatrix{0\\0\\1}$ and the world pin axis describes the rotated vector $\hat{z}_{\rm pin}^{\{0\}} = \pmatrix{ \vdots }$
The orientation of each body frame is described by the basis-vectors arranged in a 3×3 rotation matrix ${\rm R}$.
- The orientation of the parent frame described in the inertial frame is ${\rm R}_{\rm parent}^{\{0\}}$
- The relative orientation of the body frame in the parent frame is $${\rm R}_{\rm body}^{\{1\}} = {\rm rot}( \hat{z}_{\rm pin}^{\{1\}}, \theta)$$
- The orientation of the body frame described in the inertia frame is the sequence $${\rm R}_{\rm body}^{\{0\}} = {\rm R}_{\rm parent}^{\{0\}} {\rm R}_{\rm body}^{\{1\}} \tag{2}$$
I hope everything is clear so far.
Now to get to the rotational kinematics, one needs to the take the time derivative of (2). I am going to skip the derivation of the derivative on a rotating frame, and just remind you that the time derivative of an orientation ${\rm R}$ that is moving with $\vec{\Omega}$ is given by $$ \dot{\rm R} = [\vec{\Omega}\times] {\rm R} \tag{3}$$ where $[\vec{\Omega}\times] = \begin{bmatrix} 0 & -\Omega_z & \Omega_y \\ \Omega_z & 0 & -\Omega_x \\ -\Omega_y & \Omega_x & 0 \end{bmatrix}$ is the 3×3 skew-symmetric matrix that converts the cross product into linear algebra (call it the cross product matrix).
The product rule gives us the rate of rotation $$\dot{\rm R}_{\rm body}^{\{0\}} = \dot{\rm R}_{\rm parent}^{\{0\}} {\rm R}_{\rm body}^{\{1\}} + {\rm R}_{\rm parent}^{\{0\}} \dot{\rm R}_{\rm body}^{\{1\}} \tag{4}$$
- The change of the orientation frame of the parent frame described in the inertial frame is $$\dot{\rm R}_{\rm parent}^{\{0\}} = [\vec{\omega}_{\rm parent}^{\{0\}} \times] {\rm R}_{\rm parent}^{\{0\}}$$
- The change in relative orientation of the body frame in the parent frame is $$\dot{\rm R}_{\rm body}^{\{1\}} = [\hat{z}_{\rm pin}^{\{1\}} \dot{\theta}] \times {\rm R}_{\rm body}^{\{1\}}$$
- The change in orientation of the body frame described in the inertia frame is the sequence $$\dot{\rm R}_{\rm body}^{\{0\}} = [\vec{\omega}_{\rm body}^{\{0\}} \times] {\rm R}_{\rm body}^{\{0\}}$$
use the above in (4) and factor using (2) to get
$$\begin{aligned}[][\vec{\omega}_{{\rm body}}^{\{0\}}\times]{\rm R}_{{\rm body}}^{\{0\}} & =[\vec{\omega}_{{\rm parent}}^{\{0\}}\times]{\rm R}_{{\rm parent}}^{\{0\}}{\rm R}_{{\rm body}}^{\{1\}}+{\rm R}_{{\rm parent}}^{\{0\}}[(\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta})\times]{\rm R}_{{\rm body}}^{\{1\}}\\
& =[\vec{\omega}_{{\rm parent}}^{\{0\}}\times]{\rm R}_{{\rm body}}^{\{0\}}+{\rm R}_{{\rm parent}}^{\{0\}}\left([\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta}]\times{\rm R}_{{\rm body}}^{\{1\}}\right)\\
& =[\vec{\omega}_{{\rm parent}}^{\{0\}}\times]{\rm R}_{{\rm body}}^{\{0\}}+[{\rm R}_{{\rm parent}}^{\{0\}}\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta}]\times{\rm R}_{{\rm parent}}^{\{0\}}{\rm R}_{{\rm body}}^{\{1\}}\\
& =[\vec{\omega}_{{\rm parent}}^{\{0\}}\times]{\rm R}_{{\rm body}}^{\{0\}}+[{\rm R}_{{\rm parent}}^{\{0\}}\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta}]\times{\rm R}_{{\rm body}}^{\{0\}}\\
& =[\left(\vec{\omega}_{{\rm parent}}^{\{0\}}+{\rm R}_{{\rm parent}}^{\{0\}}\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta}\right)\times]{\rm R}_{{\rm body}}^{\{0\}}
\end{aligned}$$
Note that for rotation ${\rm R}$ and two vectors $a$ and $b$ you have the identity ${\rm R}(a \times b) = ({\rm R} a) \times ({\rm R} b)$
from which extract $\vec{\omega}_{{\rm body}}^{\{0\}}=\vec{\omega}_{{\rm parent}}^{\{0\}}+{\rm R}_{{\rm parent}}^{\{0\}}\hat{z}_{{\rm pin}}^{\{1\}}\dot{\theta}$ or expressed in the inertial frame
$$\vec{\omega}_{{\rm body}}^{\{0\}}=\vec{\omega}_{{\rm parent}}^{\{0\}}+\underbrace{\hat{z}_{{\rm pin}}^{\{0\}}\dot{\theta}}_{\vec{\omega}_{{\rm rel}}^{\{0\}}} \tag{5}$$
We have just derived the rotational kinematics of two connected bodies based on the time derivative of the sequence of rotations.
In summary the algebraic addition of rotational velocities comes from the multiplication of a sequence of rotations and the calculus product rule.
You can generalized the above, give a sequence of rotations ${\rm R} = {\rm R}_1 {\rm R}_2 {\rm R}_3 \ldots$ each with their own axis $\hat{z}_1$, $\hat{z}_2$, etc and angles $\theta_1$, $\theta_2$, etc.
The resulting rotational velocity of the last body is evaluated using
$$ \vec{\Omega} = \hat{z}_1 \dot{\theta}_1 + {\rm R}_1 \left( \hat{z}_2 \dot{\theta}_2 + {\rm R}_2 \left( \hat{z}_3 \dot{\theta}_3 + \ldots \right) \right) $$
And this is how you build the 3×n Jacobian matrix ${\bf J}$ such that $\vec{\Omega} = {\bf J} \pmatrix{ \dot{\theta}_1 \\ \dot{\theta}_2 \\ \vdots }$
