= Solution
Let $\sigma$ denote <surface measure on a sphere> on $S^2$ and $\nu$ volume measure on the unit ball. The <Poisson mapping theorem> can be proved directly here: counts of image points in disjoint sets are counts of the original process in their disjoint preimages. They are therefore <independent> Poisson variables, and their means are the volumes of those preimages.
For the radius map, a <measurable> set $A\subset[0,1]$ has mean
$$
\boxed{\nu_1(A)=\int_A4\pi r^2\,dr,\qquad
\nu_1([0,a])=\frac{4\pi a^3}{3}\quad(0\le a\le1).}
$$
This follows from <spherical coordinates> or by subtracting concentric-ball volumes. The disjoint-preimage argument proves that the radius image is a <Poisson process> with this <mean measure of a point process>. There is no mass at zero or at one.
The original process has no point at the origin <almost surely>, since its mean count there is zero. Thus division by $r$ in the direction map is well-defined <almost surely>. For a <measurable> surface set $C\subset S^2$, <spherical coordinates> give
$$
\boxed{\nu_2(C)=\int_C\int_0^1r^2\,dr\,\sigma(d\omega)=\frac{\sigma(C)}3.}
$$
Again disjoint surface sets have disjoint cone preimages, giving the required <independent> Poisson counts. Both <pushforward measures> are diffuse, so there are <almost surely> no image collisions; they describe the indicated point sets as well as <counting measures>.
\b[The two image processes are not <independent>.] They have exactly the same total number of points,
$$
N=\Pi_1([0,1])=\Pi_2(S^2)=\Pi(B)\sim\operatorname{Pois}(4\pi/3).
$$
Indeed,
$$
\mathbb P(\Pi_1=\varnothing,\Pi_2=\varnothing)=e^{-4\pi/3}
\ne e^{-8\pi/3}
=\mathbb P(\Pi_1=\varnothing)\mathbb P(\Pi_2=\varnothing).
$$
This alone disproves process <independence>.
More generally, split the original points into those whose radius lies in $A$ and direction lies in $C$, those satisfying only the first condition, and those satisfying only the second. Counts in these three disjoint regions are <independent>. The count shared by both images is Poisson with mean $\sigma(C)\int_A r^2dr$, so
$$
\operatorname{Cov}\bigl(\Pi_1(A),\Pi_2(C)\bigr)=\sigma(C)\int_A r^2\,dr.
$$
The fact that <Poisson polar projections share their total count> explains the dependence despite the product polar intensity $r^2dr\,\sigma(d\omega)$. Conditional on $N$, the points are <independent> uniform samples from the ball. Their radius density is $3r^2$ and their direction law is $\sigma/(4\pi)$, independently for each point. Thus the two image configurations are <independent> conditional on $N$; unconditionally the shared nonconstant count couples them.
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