Sorry I can't write in latec, so I'm adding an image here.
I don't understand how this transformation has been derived.
$$ \frac{\mathrm d}{\mathrm dt}r_e = \left(\frac{\mathrm d}{\mathrm dt}+\,\omega\,\times\right)r_i $$
where $r_e$ is position in inertial frame and $r_i$ is position in the non inertial rotating frame of reference
The quantity $\omega\times r_i$ is the tangential velocity of the rotating frame relative to the inertial frame.
$dr_i/dt$ is the velocity relative to the rotating frame, so the equation simply adds the velocity measured relative to the rotating frame to the velocity of the rotating frame relative to the inertial frame.
Let $\vec{\tilde{e_i}}$ be an orthonormal basis in the non-inertial frame and $\vec{e_i}$ an orthonormal basis in the fixed frame. I will use Einstein notation to simplify the reading.
Let $\vec{x}$ be any vector. In the fixed frame, $\vec{x} = x_i\vec{e_i}$ where the $x_i$'s are the components of $\vec{x}$ in the fixed frame. Then $\frac{d\vec{x}}{dt} = \frac{dx_i}{dt}\vec{e_i}$
In the non-inertial frame, $\vec{x} = \tilde{x_i}\vec{\tilde{e_i}}$. Then,
$\frac{d\vec{x}}{dt} = \frac{d\tilde{x_i}}{dt}\vec{\tilde{e_i}} + \tilde{x_i}\frac{d\vec{\tilde{e_i}}}{dt}$. This is is simply the chain rule, as the basis in the non-inertial frame is no longer constant.
Define $\omega_i = \mathcal{E}_{ijk} \vec{\tilde{e_j}}\cdot\frac{d\vec{\tilde{e_k}}}{dt}$. If you are not familiar with the levi-civita symbol $ (\mathcal{E}_{ijk})$, you should look it up. Let $\vec{\omega }= \omega_i \vec{\tilde{e_i}}$
Then you can show that $\tilde{x_i}\frac{d\vec{\tilde{e_i}}}{dt} = \vec{\omega} \times \vec{x}$.
So we get $\frac{d\vec{x}}{dt} = \frac{d\tilde{x_i}}{dt}\vec{\tilde{e_i}} + \vec{\omega} \times \vec{x}$.
