Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference:
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$
where $s$ denotes the space fixed frame and $r$ denotes the rotating frame.
That is the relevant equation for the instantaneous co-moving inertial frame. What about a non-comoving space frame? Generalizing this to a space frame in which the origin of the rotating frame is moving results in
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s =
\left(\frac{d\boldsymbol r_0}{dt}\right)_s
+ \left(\frac{d(\boldsymbol r - \boldsymbol r_0)}{dt}\right)_r
+\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_0)$$
where $\boldsymbol r_0$ is the displacement vector from the origin of the space frame to the origin of the rotating frame.
Suppose you use some other point $\boldsymbol r_1$ that is fixed from the perspective of the rotation frame (i.e., $\left(\frac{d(\boldsymbol r_1-\boldsymbol r_0)}{dt}\right)_r \equiv 0$. Go through the math (an exercise I'll leave up to you) and you'll find that
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s =
\left(\frac{d\boldsymbol r_1}{dt}\right)_s
+ \left(\frac{d(\boldsymbol r - \boldsymbol r_1)}{dt}\right)_r
+\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_1)$$
In other words, the angular velocity $\boldsymbol \omega$ is independent of the choice of origin.
The essence of the question is the definition of the angular velocity and its relation to the axis of rotation.
First of all, to describe the rotation of a single particle, one requires an axis of rotation $\hat{\mathbf{n}}$, and a rate of change of ‘orientation’ (or ‘angular position’) $\frac{d \theta( \hat{\mathbf{n}}) }{d t}$ around that axis. Without these 2 pieces of information, “rotation” is meaningless.
Having the axis of rotation and a rate of change of orientation, one can define the angular velocity $\vec{\omega}$ as
$$
\boldsymbol{\omega} = \frac{d \theta}{d t} \, \hat{\mathbf{n}}
$$
which includes both pieces of information.
Notice that the angular velocity is parallel to the axis of rotation by definition, and has the same amount of ‘information’ as the particle velocity $\mathbf{v}$.
To clarify this further, consider the following figure which depicts the rotation in 3d space, and conforms to the example given in the question,
where the spherical-polar coordinates $\theta$ and $\phi$ are used as the azimuthal and polar angles, respectively and the zenith ($z$-axis) is taken to be parallel to $\hat{\mathbf{n}}$.
In the limit of infinitesimal change in time, $\Delta t \rightarrow 0$,
$$
\Delta \mathbf{r} \approx r \, \sin \phi \, \Delta \theta ~,
$$
and thus,
$$
\left| \frac{d \mathbf{r}}{d t} \right| = r \sin \phi \frac{d \theta}{d t} ~.
$$
One clearly observes that both the magnitude and the direction of $\frac{d \mathbf{r}}{d t}$ (which is perpendicular to the plane defined by the particle position $\mathbf{r}$ and the rotation axis $\hat{\mathbf{n}}$) are given correctly by the cross product
$$
\frac{d \mathbf{r}}{d t} = \hat{\mathbf{n}} \times \mathbf{r} \frac{d \theta}{d t} ~.
$$
Since $\frac{d \mathbf{r}}{d t} \equiv \mathbf{v}$ and $\hat{\mathbf{n}} \frac{d \theta}{d t} \equiv \vec{\omega}$, one obtains
$$
\mathbf{v} = \frac{d \mathbf{r}}{d t} = \boldsymbol{\omega} \times \mathbf{r} ~.
$$
Therefore, the instantaneous axis of rotation $\hat{\mathbf{n}}(t)$ and the instantaneous angular velocity $\vec{\omega}(t)$ are indeed parallel.
Finally, as shown by Gary Godfrey in a comment, one can obtain the correct relation for the angular velocity $\boldsymbol{\omega}$ in terms of the velocity $\mathbf{v}$ and position $\mathbf{r}$ as
\begin{align}
\mathbf{r} \times \mathbf{v} &= \mathbf{r} \times ( \boldsymbol{\omega} \times \mathbf{r} ) \stackrel{\text{BAC-CAB}}{=} \boldsymbol{\omega} \, (\mathbf{r} \cdot \mathbf{r} ) - \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega}) \\
\Rightarrow \boldsymbol{\omega} &= \frac {(\mathbf{r} \times \mathbf{v}) + \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega})}{r^2} ~.
\end{align}
This answer is based on Kleppner, D., and R. Kolenkow. “An introduction to mechanics”, (2ed, 2014), pp. 294–295. The figure is taken from the same source.
Best Answer
Starting with the rotations matrix \begin{align*} &[\,_1^3\,\mathbf S\,]=[\,_1^2\,\mathbf S\,]\,[\,_2^3\,\mathbf S\,]\quad\Rightarrow\quad [\,_1^3\,\mathbf{\dot{S}}\,]=[\,_1^2\,\mathbf{\dot{S}}\,]\,[\,_2^3\,\mathbf S\,]+ [\,_1^2\,\mathbf S\,]\,[\,_2^3\,\mathbf{\dot{S}}\,]\\ &\text{with}\quad \mathbf{\dot{S}}=\mathbf{\tilde{\omega}}\,\mathbf S\quad \mathbf{\tilde{\omega}}= \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]\quad\Rightarrow \\\\ &\mathbf{\tilde{\omega}}_{13}[\,_1^3\,\mathbf{{S}}\,]= \mathbf{\tilde{\omega}}_{12}[\,_1^2\,\mathbf{{S}}\,]\,[\,_2^3\,\mathbf S\,]+ [\,_1^2\,\mathbf S\,]\,\mathbf{\tilde{\omega}}_{23}[\,_2^3\,\mathbf{{S}}\,]\\\\ &\text{multiply from the right with}\quad [\,_3^1\,\mathbf{{S}}\,]\\\\ &\mathbf{\tilde{\omega}}_{13}= \mathbf{\tilde{\omega}}_{12}\underbrace{[\,_1^2\,\mathbf{{S}}\,]\,[\,_2^3\,\mathbf S\,][\,_3^1\,\mathbf{{S}}\,]}_{I_3}+ [\,_1^2\,\mathbf S\,]\,\mathbf{\tilde{\omega}}_{23}\underbrace{[\,_2^3\,\mathbf{{S}}\,][\,_3^1\,\mathbf{{S}}\,]} _{ [\,_2^1\,\mathbf S\,]}\\ &\text{thus the angular velocity vector}\\ &\mathbf\omega_{13}=\mathbf\omega_{12}+[\,_1^2\,\mathbf S\,]\mathbf\omega_{23} \end{align*} \begin{align*} &\text{with}\\ &[\,_1^2\,\mathbf S\,]=\left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right] \quad, [\,_2^3\,\mathbf S\,]=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \phi \right) &-\sin \left( \phi \right) \\ 0 &\sin \left( \phi \right) &\cos \left( \phi \right) \end {array} \right] \\ &\mathbf\omega_{12}=\begin{bmatrix} 0 \\ 0 \\ \omega_\psi \\ \end{bmatrix}\quad \mathbf\omega_{23}=\begin{bmatrix} \omega_\phi \\ 0 \\ 0 \\ \end{bmatrix} \end{align*}