Just in the case $n=3$, it seems that the natural fibration $F_4S^2\to F_3S^2$ (forget the last point) has a section. The section $F_3S^2 \to F_4 S^2$ is defined by the formula $$(v_1, v_2, v_3)\mapsto (v_1, v_2, v_3, \frac{(v_2-v_1)\times (v_3 - v_1)}{||(v_2-v_1)\times (v_3 - v_1)||}).$$ Here $v_1, v_2, v_3$ are three distinct unit vectors in ${\mathbb R}^3$, representing a point in $F_3S^2$. The idea is that you consider the line through the origin perpendicular to the plane containing the points $v_1, v_2, v_3$, and take the intersection of that line with $S^2$ to be your fourth point. Use the ordering of the points to decide which of the two points to take.
Such a section defines a split injection from the third pure spherical braid group to the fourth.
It seems that
Edit: Ricardo's answer shows a simpler way to add a fourth point, and also generalizes the map does not generalize argument to produce a section of the fibration $n>3$. Nor does it work F_mS^2 \to F_nS^2$ for the full spherical braid group, even when all $n=3$.m>n \ge 3$.
