Suppose we have two moving coordinate frames $F_1$ and $F_2$. We know their position $p_{F_1}^W, p_{F_2}^W \in \mathbb{R}^3 $ as well as the respective orientations $R_{F_1}^W, R_{F_2}^W \in SO\left(3\right)$. Furthermore we know both frames' translational velocities $v_{F_1}^W, v_{F_2}^W \in \mathbb{R}^3$ and angular velocities $\omega_{F_1}^W, \omega_{F_2}^W \in \mathbb{R}^3$. All these quantities are given with respect to common reference frame $W$.
However I want to express all these quantities with respect to frame $F_1$.
What I could come up with:
$R^{F_1}_{F_2} = {R_{F_1}^{W}}^T R_{F_2}^W $ (Orientation of frame $F_2$ w.r.t $F_1$)
$p_{F_2}^{F_1} = {R_{F_1}^{W}}^T\left(p_{F_2}^W- p_{F_1}^W\right)$ (Position of frame $F_2$ w.r.t $F_1$)
$\omega_{F_1}^{F_1} = {R_{F_1}^{W}}^T\omega^W_{F_1}$ (Body fixed angular velocity of $F_1$) (Not sure)
$v_{F_1}^{F_1} = {R_{F_1}^{W}}^T v^W_{F_1}$ (Body fixed translational velocity of $F_1$) (Not sure)
Most importantly: I want to know the quantity $\omega_{F_2}^{F_1}$ and $v_{F_2}^{F_1}$ that is the angular and translational velocity of Frame $F_2$ with respect of (as measured from) Frame $F_1$.
Let me elaborate. Let $\vec A$ be an arbitrary time dependent vector (I am using here full "physics" vector notation). The derivative of this vector with respect to time in the frames $W$, $F_1$, and $F_2$ are different, and related to each other by:
$$ \frac{d^{F_1}\vec A}{dt}= \frac{d^{W}\vec A}{dt}-\vec\omega_{F_1}^W\!\times\!\vec A,\tag{1}$$ $$ \frac{d^{F_2}\vec A}{dt}= \frac{d^{W}\vec A}{dt}-\vec\omega_{F_2}^W\!\times\!\vec A.$$
This two equations, by the way, define $\vec\omega_{F_1}^W$ and $\vec\omega_{F_2}^W$. Combining the two we get:
$$ \frac{d^{F_2}\vec A}{dt}= \frac{d^{F_1}\vec A}{dt}-(\vec\omega_{F_2}^W-\vec\omega_{F_1}^W)\!\times\!\vec A.$$
Because $\vec A$ is arbitrary, this shows that:
$$\bbox[5px,border:2px solid red]{\vec\omega_{F_2}^{F_1}=\vec\omega_{F_2}^W-\vec\omega_{F_1}^W }$$
To get the second equation, we can use the definitions of $\vec v_{F_1}^W$, $\vec v_{F_2}^W$ and $\vec v_{F_2}^{F_1}$:
$$\vec v_{F_1}^W = \frac{d^W \vec p_{F_1}^W}{dt},$$ $$\vec v_{F_2}^W = \frac{d^W \vec p_{F_2}^W}{dt},$$ $$\vec v_{F_2}^{F_1} = \frac{d^{F_1}\vec p_{F_2}^{F_1}}{dt}.$$
But $\vec p_{F_2}^{F_1}=\vec p_{F_2}^W-\vec p_{F_1}^W$, so using equation $(1)$ we get:
$$\bbox[5px,border:2px solid red]{\vec v_{F_2}^{F_1} = \vec v_{F_2}^W-\vec v_{F_1}^W -\vec\omega_{F_1}^W\!\times\!(\vec p_{F_2}^W-\vec p_{F_1}^W)}$$