Let us suppose we have two orthogonal rotation matrices representing a three-dimensional rotations $\mathbf{R}(t)$ and $\mathbf{R}(t+\Delta t)$
How is it possible to extract the angular velocity of the rotation $\boldsymbol \omega \in \mathbb{R}^3$ or equivalently the angular velocity tensor (represented by the skew-symmetric matrix) $\boldsymbol \Omega = \begin{pmatrix} 0 & -\omega_z(t) & \omega_y(t) \\ \omega_z(t) & 0 & -\omega_x(t) \\ -\omega_y(t) & \omega_x(t) & 0 \\ \end{pmatrix}$ from them?
I would ask you if the following approach make sense...
1) By mean of exponential map $\mathbf{R}(t+\Delta t) = e^{\boldsymbol \Omega \Delta t} \mathbf{R}(t)$
2) Solve for $\boldsymbol \Omega$ then $\boldsymbol \Omega = \frac{ \log\left( \mathbf{R}(t)\mathbf{R}^{-1}(t+\Delta t) \right)}{\Delta t}$
3) Call $\boldsymbol Y = \mathbf{R}(t) \mathbf{R}^{-1}(t+\Delta t) -\mathbf{I}$ and approximate the logarithm with its Taylor expansion
$ \boldsymbol \Omega(t) = \dfrac{ \log\left(\mathbf{R}(t) \mathbf{R}^{-1}(t+\Delta t) \right)}{\Delta t} \approx \frac{1}{\Delta t} \left( \mathbf{Y} - \frac{\mathbf{Y}^2}{2} + \frac{\mathbf{Y}^3}{3} - \frac{\mathbf{Y}^4}{4} + \ldots \right ) $