Let $f,g \in PSL(2,R)$ be isometries of $\mathbb{H}^2$ of hyperbolic type. Let $h=[f,g]$ be their commutator.
Is there an explicit geometric criterion to determine if $h$ is hyperbolic, parabolic or elliptic?
With "explicit geometric criterion" I mean something like this: $h$ is the identity iff $f$ and $g$ have the same fixed points on $\partial \mathbb{H}^2$.
There is a section “The Geometry of Commutators” on pg 184 in Alan Beardon’s book “The Geometry of Discrete Groups” that investigates just this kind of thing. This kind of reasoning can be found in that section:
Suppose both $f,g$ are hyperbolic and the axis of $f$ and $g(\mbox{axis of f})$ do not meet. To classify $[f,g]$ as elliptic, parabolic or hyperbolic, work in the disc model and consider $[f,g]$ as the composition of $f$ with $gf^{-1} g^{-1}$. Notice that the axis of $gf^{-1} g^{-1} =g(\mbox{axis of f})$.
Now write $f$ and $gf^{-1} g^{-1}$ as the composition of two reflections:
$$ f=ab \ \ \ \ \mbox{and} \ \ \ \ gf^{-1} g^{-1}=bc$$
where for convenience $b$ is chosen to be the reflection in the real axis. We can do this as any hyperbolic element in $PSL(2,\mathbb{R})$ can be written as the composition of two reflections in geodesics perpendicular to its axis, either of which we may choose. Since the axes of $f$ and $gf^{-1} g^{-1}$ don't meet they have a common perpendicular which by conjugation we can take to be the real line.
Using the same name for both the reflection and the line in which it reflects, the axis of $f$ is orthogonal to the both $a$ and $b$, while the axis of $gf^{-1} g^{-1}$ that is $g(\mbox{axis of f})$, is orthogonal to both $b$ and $c$.
Now $[f,g]=ac$ may be classified according to if and where the lines $a$ and $c$ meet. If they meet inside the disc this meeting point is a fixed point of the commutator which must be elliptic. If $a$ and $c$ land at a common point on the unit circle, this is the only fixed point of the commutator which must therefore be parabolic. If the lines don’t meet either in the unit disc or its boundary then the commutator is hyperbolic.
It turns out the trace of the commutator is well defined, that is either representative of $f$ or $g$ in $SL(2,\mathbb{C})$ yields the same value for trace$[f,g]$. With this in mind, the following (see Beardon Thm 4.3.5) addresses your last sentence, and is true of all maps $f,g \in PSL(2,\mathbb{C})$:
(i) The maps $f$,$g$ have a common fixed point if and only if trace$[f,g]=2$
(ii) If both $f$ and $g$ are not the identity and have a common fixed point in $\widehat{\mathbb{C}}$ then either:
(a) $[f,g]=I$ and the set of fixed points of $f$ equals the set of fixed points of $g$; or
(b) $[f,g]$ is parabolic and the sets of fixed points above are not equal.