Recall the expression for $d\rr$ in polar coordinates, namely $$d\rr = dr\,\rhat + r\,d\phi\,\phat$$ “Dividing” this expression by $dt$ yields an expression for the velocity in polar coordinates which you may already have seen, namely $$\dot{\rr} = \dot{r}\,\rhat + r\,\dot\phi\,\phat$$ But recall that the position vector in polar coordinates takes the form $$\rr = r\,\rhat$$ and we can differentiate this directly to obtain $$\dot{\rr} = \dot{r}\,\rhat + r\,\rhatdot$$ Comparing these two expressions for $\dot{\rr}$ yields $$\rhatdot = \dot\phi\,\phat$$

Assume now that we are in a rotating frame, with constant angular velocity $\Om=\Omega\,\zhat$. Then $\phi=\Omega\,t$, and we have $$\rhatdot = \Omega\,\phat = \Omega\,\zhat\times\rhat = \Om\times\rhat $$ Differentiating this expression again yields $$\Omega\,\phatdot = \Om\times\rhatdot = \Om\times\Omega\,\phat $$ since $\Om$ is constant, and dividing by $\Omega$ now yields $$\phatdot = \Om\times\phat$$

Thus, for any vector $\ww=w_r\,\rhat+w_\phi\,\phat$, we have $$\dot{\ww} = \dot{\ww}_{\!R} + \Om\times\ww$$ where we have written $$\dot{\ww}_{\!R} = \dot{w}_r\,\rhat+\dot{w}_\phi\,\phat$$ for the “naive” derivative, ignoring the time-dependence of the rotating basis vectors. But it is precisely this naive derivative which would be computed by a rotating observer!

The rest is easy. The velocity computed by a rotating observer would be $$\vv_{\!R} = \vv - \Om\times\rr$$ so that the effective acceleration would be \begin{eqnarray} \aa_{\!R} &=& \left(\vv - \Om\times\rr\right)^\cdot - \Om\times \left(\vv - \Om\times\rr\right) \\ &=& \aa - 2\,\Om\times\vv + \Om\times(\Om\times\rr) \\ &=& \aa - 2\,\Om\times\vv_{\!R} - \Om\times(\Om\times\rr) \end{eqnarray}