There is a very physical picture to this, if you are willing to work with the disk model of hyperbolic space, instead of the upper half plane, to which it is related by an isometry.

The Lie group $\mathrm{SL}(2,\mathbb{R})$ is a double cover of the identity component $\mathrm{SO}_0(2,1)$ of $\mathrm{O}(2,1)$, which is the Lorentz group in 3 dimensions.  In other words, $\mathrm{O}(2,1)$ is the subgroup of $\mathrm{GL}(3,\mathbb{R})$ which preserves a symmetric inner product $\eta$ of signature $(2,1)$:
$$\eta = \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 \end{pmatrix}.$$

Now consider the two-sheeted hyperboloid in $\mathbb{R}^3$ defined by $x^2 + y^2 - z^2 = -1$.  The upper sheet -- let's call it $\mathbb{D}$ -- with $z>0$ is topologically a disk.  It inherits a riemannian metric from the Minkowski metric on $\mathbb{R}^3$ defined by $\eta$.
The identity component of $\mathrm{O}(2,1)$ acts on $\mathbb{D}$ as isometries.

The isotropy at the point $(0,0,1)$ consists of rotations in the $x,y$-plane, whence it is isomorphic to $\mathrm{SO}(2)$.  Hence $\mathbb{D} = \mathrm{SO}_0(2,1)/\mathrm{SO}(2)$.

Notice that it is is $\mathrm{SO}_0(2,1)$ (a.k.a. $\mathrm{PSL}(2,\mathbb{R})$) which acts effectively on $\mathbb{D}$ and not $\mathrm{SL}(2,\mathbb{R})$.


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**Added** I forgot to relate the disk to the upper half plane.  If you think of $\mathbb{D}$ as the unit disk in the complex plane, then the map $\mathbb{D} \to \mathbb{H}$ is given by the following Möbius transformation:
$$ z \mapsto \frac{z-i}{z+i}$$