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
Let's first start with a picture. The object (blue ball) is rotating around the AB axis with angular velocity $\vec\omega_0$.
Next, the $AB$ axis is rotating around the vertical with an angular velocity $\vec \omega '$. We are told that $\omega'$ is constantly being accelerated by an angular acceleration $\alpha$, and from our kinematic equations (recall $\displaystyle \alpha = \dfrac{d\omega'}{dt} \implies \omega' = \int\alpha \, \mathrm dt = \alpha t $) we find that $\vec \omega ' = \alpha t \hat z$.
We need to also find out what $\vec \omega_0$ is -- and let's work in cylindricalcoordinates here -- since the $AB$ axis is always rotating in the $xy$ plane, we can write $\vec\omega_0 = \omega_0 \hat r$ where $\hat r$ is the radial unit vector pointing along $\vec\omega_0$.
The angular acceleration of the body is \begin{align} \dfrac{d\vec\omega_0}{dt} &= \dfrac d{dt} \left( \omega_0\hat r \right)\\ &= \omega_0\dfrac{d\hat r}{dt} \end{align}
Using the cartesian basis, one can show that $\dfrac{d\hat r}{dt}=\omega' \hat \varphi$ to get $$\dfrac{d\vec\omega_0}{dt} = \omega_0 \omega'\hat\varphi.$$
Next, recall in cylindrical coordinates that $\hat\varphi = \hat z \times \hat r$ so we can substitute that in and get $$\dfrac{d\vec\omega_0}{dt} = \omega' \hat z\times \omega_0\hat r$$ which is exactly $\vec \omega' \times \vec \omega_0$.