= Solution
Apply part (a) first to the <position> $\mathbf r$:
$$
(\dot{\mathbf r})_S=(\dot{\mathbf r})_{S'}+\boldsymbol\omega\times\mathbf r.
$$
Applying it again to both terms on the right gives
$$
(\ddot{\mathbf r})_S=(\ddot{\mathbf r})_{S'}+\boldsymbol\omega\times(\dot{\mathbf r})_{S'}+(\dot{\boldsymbol\omega})_{S'}\times\mathbf r+\boldsymbol\omega\times(\dot{\mathbf r})_S,
$$
where $(\dot{\boldsymbol\omega})_S=(\dot{\boldsymbol\omega})_{S'}$ because $\boldsymbol\omega\times\boldsymbol\omega=0$. Substituting the first formula into the last term produces
$$
\boxed{(\ddot{\mathbf r})_S=(\ddot{\mathbf r})_{S'}+2\boldsymbol\omega\times(\dot{\mathbf r})_{S'}+(\dot{\boldsymbol\omega})_{S'}\times\mathbf r+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)}.
$$
In the Earth-fixed <rotating reference frame>, the sideways apparent force is the <Coriolis acceleration> multiplied by the ski-doo's <mass>:
$$
\mathbf F_{\rm side}=-2m\boldsymbol\omega\times\mathbf v.
$$
At the South Pole, $\boldsymbol\omega$ is vertical and antiparallel to the local outward normal, while $\mathbf v$ is horizontal, so
$$
\boxed{|\mathbf F_{\rm side}|=2m\omega v}.
$$
The force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.
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