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

Assume now that we are in a rotating frame, with constant angular velocity $$\Om=\Omega\,\zhat=\Omega\,(\cos\theta\,\rhat-\sin\theta\,\that)$$ Then $\dot\theta=0$ and $\dot\phi=\Omega$, and we have $$\rhatdot = \Omega\sin\theta\,\phat = \Omega\,\zhat\times\rhat = \Om\times\rhat $$ Differentiating this expression again yields $$\Omega\sin\theta\,\phatdot = \Om\times\rhatdot = \Om\times\Omega\sin\theta\,\phat $$ since $\Om$ is constant, and dividing by $\Omega\sin\theta$ now yields $$\phatdot = \Om\times\phat$$

Finally, consider the derivative of $\that=\phat\times\rhat$. We have $$\thatdot = \left(\phat\times\rhat\right)^\cdot = \phatdot\times\rhat + \phat\times\rhatdot = (\Om\times\phat)\times\rhat + \phat\times(\Om\times\rhat) $$ The last term is zero, since $\rhatdot$ is parallel to $\phat$, so that we have: 1) $$\thatdot = (\Om\times\phat)\times\rhat = \Om\times(\phat\times\rhat) = \Om\times\that $$

Thus, for any vector $\ww=w_r\,\rhat+w_\theta\,\that+w_\phi\,\phat$, we have $$\dot{\ww} = \dot{\ww}_{\!R} + \Om\times\ww$$ where we have written $\dot{\ww}_{\!R} = \dot{w}_r\,\rhat+\dot{w}_\theta\,\that+\dot{w}_\phi\,\phat$ 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. Just as in the 2-dimensional case, the velocity and acceleration computed by a rotating observer are: \begin{eqnarray} \vv_{\!R} &=& \vv - \Om\times\rr \\ \aa_{\!R} &=& \aa - 2\,\Om\times\vv_{\!R} - \Om\times(\Om\times\rr) \end{eqnarray}