To get the angular velocity from a time-dependent rotation matrix $R(t)\in \mathrm{SO}(3)$, all you need to do is differentiate.
More specifically, since you know that $R(t)^TR(t)\equiv \mathbb 1$, differentiating with respect to time you get $R(t)^TR'(t)+R'(t)^TR(t) = 0$, and w.l.o.g. taking $t=0$ and therefore $R(t)=\mathbb 1$, you get $R'(t) = -R'(t)^T$, i.e. that the derivative of the rotation matrix is skew symmetric. This means that you can write it as
$$
\frac{\mathrm dR}{\mathrm dt} = \begin{pmatrix}0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0\end{pmatrix}
$$
for some triplet of numbers $\omega_x,\omega_y,\omega_z$. These turn out to be the components of the angular velocity, at least at those times when $R(t)=\mathbb 1$. In general you need to do some playing around with frames of reference, which are duly explained in Wikipedia, but the idea remains unchanged.
The mechanism you propose, i.e. getting the angular velocity directly from the Euler angles, is rather more complicated - for the details see a previous question on this site, Angular Velocity expressed via Euler Angles.
There are two ways to look at linear acceleration of a rigid body.
- Material Acceleration You can look at the acceleration of a specific particle on the body in terms of how the linear velocity components change over time. We call this the material acceleration. So we are are tracking a point A that is riding on the rotating reference frame
$$\overline{a}_A^m = \tfrac{\rm d}{{\rm d}t} \overline{v}_A \tag{1}$$
- Spatial Acceleration You can also look at the change in speed of whatever particle is passing under a fixed reference point
$$\overline{a}_A^s = \tfrac{\partial}{\partial t} \overline{v}_A \tag{2}$$
See spatial acceleration entry on wikipedia.
The two are related with a familiar relationship
$$ \boxed{ \overline{a}_A^m = \overline{a}_A^s + \overline{\omega} \times \overline{v}_A } \tag{3}$$
This is a direct result of the formula for differentiating a vector $\overline{A}$ on rotating frame
$$ \tfrac{\rm d}{{\rm d}t} \overline{A} = \tfrac{\partial }{\partial t} \overline{A} + \overline{\omega} \times \overline{A} \tag{4}$$
See rotating frames on MIT lecture series
Looking at your notation $a_b$ is a spatial acceleration and $a_b + \omega \times v_b$ the material acceleration.
Take the simple example of a spinning disk with non-zero rotational acceleration.
Look at point A on the disc located at $\vec{r}_A$ from the center at some point the velocity vector is
$$ \overline{v}_A = \overline{\omega} \times \overline{r} \tag{5}$$
Looking at the material acceleration the point is moving with time
$$ \overline{a}_A^m = (\tfrac{\rm d}{{\rm d}t} \overline{\omega} ) \times \overline{r}_A + \overline{\omega} \times \tfrac{\rm d}{{\rm d}t} \overline{r}_A = \overline{\alpha} \times \overline{r}_A + \overline{\omega} \times \overline{v}_A \tag{6}$$
But the spatial acceleration where the point is fixed in space
$$\require{cancel} \overline{a}_A^s = (\tfrac{\rm d}{{\rm d}t} \overline{\omega} ) \times \overline{r}_A + \overline{\omega} \times \cancel{ \tfrac{\rm d}{{\rm d}t} \overline{r}_A }= \overline{\alpha} \times \overline{r}_A \tag{7}$$
As you can see the spatial acceleration (which is sometimes called Euler acceleration) contains only the result of the change in rotational speed of the disk. This is why it is useful because it describes the acceleration state of the entire body and not only a specific point in space.
You can transfer spatial acceleration from one point to another (say B) just as you do velocities
$$ \begin{aligned}
\overline{v}_B & = \overline{v}_A + \overline{\omega} \times ( \overline{r}_B - \overline{r}_A) \\
\overline{a}_B^s & = \overline{a}_A^s + \overline{\alpha} \times ( \overline{r}_B - \overline{r}_A) \end{aligned} \tag{8} $$
and then convert to material acceleration by (3)
$$ \overline{a}_B^m = \overline{a}_B^s + \overline{\omega} \times \overline{v}_B \tag{9}$$
Best Answer
I like the equation in the following form, all expressed on the same basis vectors.
$$ \dot{R} = [\omega \times] R $$
This is used to derive the kinematics of joints where $R = R_1 R_2 \ldots $
Now you can definitely take the time derivative of the above using the chain rule to get
$$ \ddot{R} = [\dot{\omega} \times] R + [\omega \times] \dot{R} = [\dot{\omega} \times] R + [\omega \times] [\omega\times] R $$
or
$$ [\dot{\omega} \times] = \ddot{R} R^\top - [\omega \times] [ \omega\times] R R^\top = \ddot{R} R^\top - [\omega \times] [\omega\times] $$
Notice that $[\omega \times] [\omega\times]$ is a symmetric matrix and $\ddot{R} R^\top$ a general matrix. If the above is valid this means that $\ddot{R} R^\top$ is split into symmetric and anti symmetric parts as follows
$$ \ddot{R} R^\top = \underbrace{[\dot{\omega}\times]}_{\text{skew}} + \underbrace{ [\omega \times][\omega \times]}_{\text{symm}} $$