A good way to see this is by thinking of Galois conjugation. Many discrete groups acting on the hyperbolic plane $H^2$ (whose orientation preserving isometry group is $PSL(2,\mathbb R)$ or in hyperbolic 3-space (with isometry group $PSL(2,\mathbb C)$) have Galois conjugates in $PSU(2) = SO(3)$, acting on $S^2$.

It's very easy to geometrically establish certain subgroups of isometries of $H^2$ or $H^3$ are free:
in the plane for any 4 disjoint half-planes $U,X,Y,Z$, isometries $A$ sending the complement of $U$ to $X$ and $B$ sending the complement of $Y$ to $Z$ generate a free group. For generic $A$ and $B$, they have enough algebraic independence that they are Galois conjugate to a pair of elements in $PSU(2)$, therefore giving a free subgroup.

**Edited to correct error noted in comment**: A particular example that is easy to see algebraically before understanding things in greater generality is the $(777)$ triangle group generated by two $2\pi/ 7$ rotations in $H^2$ such that there product is also a $2 \pi/7$
$\left < a, b, c | abc=a^7=b^7=c^7 = 1 \right > $, it has 2 Galois conjugates within
$PSL(2,\mathbb C)$, where the rotations become $2/7*2\pi$ and $3/7*2\pi$. The second of these is in $PSU(2) = SO(3)$ and acts on $S^2$: this Galois conjugate comes from a spherical triangle with three angles of
$3/7 \pi$: you rotate about the corners of the triangles by twice the angle of the triangle.

The Galois action can be shown in elementary terms for anyone who understands that the different primitive 7th roots of unity are algebraically isomorphic (Galois conjugate): in
$SL(2,\mathbb C)$, the trace of any element in a 2-generator group is determined by the traces
of $a$, $b$, and $ab$, using the trace relation $T(a*b) + T(a*b^{-1}) = T(a)T(b)$, which is a simple consequence of the fact that a matrix satisfies its characteristic polynomial.
Traces of other elements in the group are polynomials in these three traces, so they're determined by the algebraic relationships in this ring.
From this it follows it follows that this $(777)$ triangle group acts faithfully on $S^2$ (taking into account that $SO(3) = SU(2)/\pm 1$ and $SU(2) \subset SL(2,\mathbb C)$.)

It's easy to write down free subgroups of any hyperbolic group like $777$: there are simple geometric sufficient criterion by constructing fundamental domains. These turn into free subgroups of $SO(3)$.

For the other example mentioned, generated by $A$ and $B$, if we lift to $SL(2,\mathbb C)$ and choose the traces of $A$, $B$ and $AB$ to be algebraically independent, then we can map them to 3 transcendentally independent elements in the interval $(-2,2)$ to conjugate the group inside an $SU(2)$. Or, we can do the same thing by choosing appropriate elements within any number field except the rationals and the quadratic imaginary field. (It's also straightforward to get free subgroups of $SO(2, \mathbb Q)$, by using a different geometric picture).

Nearly any subgroup of $SO(3)$ has at least one galois conjugate that has unbounded orbits in $PSL(2,\mathbb C)$. In any such case, you can find a free subgroup geometrically by picking elements that give a good fundamental domain.