Let us actually discuss more generally, because your picture and argument are very limited to 3D (especially with the invocation of cross products), and make too much appeal to vague intuition (which I had great trouble ‘getting’ as a student).
The general setup for discussing a particle moving in $n$ dimensions $\Bbb{R}^n$, under pure rotation is that the particle starts off at some initial point $\mathbf{r}_0\in\Bbb{R}^n$, and its position at some later time $t$ is given by
\begin{align}
\mathbf{r}(t)=R(t)\cdot \mathbf{r}_0,\tag{$*$}
\end{align}
where $R(t)$ is an $n\times n$ rotation matrix, meaning it satisfies
\begin{align}
R(t)\cdot R(t)^{\top}=R(t)^{\top}R(t)=I\tag{$**$}
\end{align}
(and the extra condition of $\det R(t)=+1$, but let’s ignore this condition). The condition $(**)$ expresses that the matrix $R(t)$ when applied to any vector, keeps its length fixed (i.e $R(t)\cdot \xi$ and $\xi$ have the same length for all vectors $\xi$), and since the lengths stay the same, this means the ‘angles’ that $\xi$ and $R(t)\cdot \xi$ make must be the only difference, hence the name rotation matrix (actually reflections also preserve lengths, and that’s why there’s the extra condition of positive determinant). Also, we immediately read off that the inverse is the transpose: $R(t)^{-1}=R(t)^{\top}$.
Anyway, with that brief intro to rotation matrices out of the way, we can proceed very easily. Let us now find the velocity of the particle undergoing pure rotation, i.e we differentiate both sides of the equation $\mathbf{r}(t)=R(t)\cdot\mathbf{r}_0$ to get
\begin{align}
\dot{\mathbf{r}}(t)&=\dot{R}(t)\cdot\mathbf{r}_0\tag{$***$}
\end{align}
But now, we can rewrite out original equation $(*)$ as $\mathbf{r}_0=R(t)^{-1}\cdot \mathbf{r}(t)=R(t)^{\top}\cdot\mathbf{r}(t)$ (the inverse matrix is the transpose as mentioned above). So, plugging this into $(***)$, we get
\begin{align}
\dot{\mathbf{r}}(t)&=\dot{R}(t)R(t)^{\top}\cdot \mathbf{r}(t)\equiv \Omega(t)\cdot\mathbf{r}(t),
\end{align}
where we have defined $\Omega(t)=\dot{R}(t)R(t)^{\top}$. One very important feature of this matrix $\Omega(t)$ is that it is skew-symmetric. The way we see this is by differentiating both sides of $R(t)R(t)^{\top}=I$ in $(**)$. This implies by the product rule, $\dot{R}(t)R(t)^{\top}+R(t)\dot{R}(t)^{\top}=0$, i.e $\Omega(t)+\Omega(t)^{\top}=0$, which says exactly that $\Omega(t)$ is skew-symmetric.
Hence, we get our desired equation $\dot{\mathbf{r}}(t)=\Omega(t)\cdot\mathbf{r}(t)$, which says the velocity of the particle equals some skew-symmetric matrix multiplied by the position of the particle. Finally, in the special case of $n=3$ dimensions, skew-symmetry means we can write it as
\begin{align}
\Omega(t)&=
\begin{pmatrix}
0&a(t)&b(t)\\
-a(t)&0& c(t)\\
-b(t)&-c(t)&0
\end{pmatrix},
\end{align}
for some functions $a(t),b(t),c(t)$. However, at this stage, it is tradition to write $a(t)=-\omega_3(t), b(t)=\omega_2(t),c(t)=-\omega_1(t)$, so
\begin{align}
\Omega(t)&=
\begin{pmatrix}
0&-\omega_3(t)&\omega_2(t)\\
\omega_3(t)&0& -\omega_1(t)\\
-\omega_2(t)&\omega_1(t)&0
\end{pmatrix}.
\end{align}
The reason for this funny tradition of naming is that if you now carry out the matrix multiplication for the case $n=3$ in $\dot{\mathbf{r}}(t)=\Omega(t)\cdot\mathbf{r}(t)$, then you get exactly the desired equation
\begin{align}
\dot{\mathbf{r}}(t)=\mathbf{\omega}(t)\times \mathbf{r}(t).\tag{@}
\end{align}
Terminology:
- $R(t)$ is sometimes called the development of the path $\mathbf{r}(t)$ in the rotation group.
- $\Omega(t)$ is called the angular velocity matrix/operator.
- In $n=3$ dimensions, $\omega(t)\in\Bbb{R}^3$ is called the angular velocity vector (associated with $R(t)$, or with $\Omega(t)$).
Edit:
To be super explicit, given this definition of $\Omega(t)$, we can define $\Theta(t)$ to be an anti-derivative (which is unique up to an additive constant determined by initial conditions (which we may WLOG set to $0$)). Then, by definition, we will have $\dot{\Theta}(t)=\Omega(t)$, and hence $\dot{\mathbf{r}}(t)=\dot{\Theta}(t)\cdot\mathbf{r}(t)$ ($\cdot$ is the multiplication between a matrix and a column vector).
In the case $n=3$, we can of course identify these skew-symmetric matrices with vectors as described above, so from $\omega(t)$, we get a primitive $\theta(t)$ (which to be super explicit, is NOT just the 3 Euler angles stuck together in a column vector) such that $\dot{\mathbf{r}}(t)=\dot{\theta}(t)\times\mathbf{r}(t)$.
