Exactly horizontal geodesics (perpendicular to the isotropy subgroup of the point in hyperbolic space) project to geodesics. Others do not, they give rise to more general curves
(called ballistic curves here).

# Edit (twice)

To be specific, my remark applies to the situation that $pr:SL_2(\mathbb R)\to \mathbb H^2$
(given by acting on $i\in \mathbb H^2$)
is a Riemannian submersion. Since we have a left action on $\mathbb H^2$, a suitable right invariant metric on $SL_2$ would furnish this. In such a situation one has: a geodesic in $SL_2$ which is perpendicular to the fibers of $pr$ at one point, is perpendicular everywhere and it projects to a geodesic on $\mathbb H^2$.

You look at a left invariant metric, but one which is given by an $SO_2$-invariant inner product on the Lie algebra. Inversion on $SL_2$ is an isometry from right to left invariant metric fixing the isotropy $SO_2$ of $i\in \mathbb H^2$. The right action of $SO_2$ on $SL_2$ is thus by isometries for the left invariant metric you use.

In this case, let me describe the geodesic equation on $SL_2$ (which differs just by a sign from the better known right invariant case)
A smooth curve $t \mapsto g(t)\in SL_2$ is a geodesic iff
$$a(t) = g(t)^{-1}.g'(t) \in \mathfrak{sl}_2$$
satisfies
$$\partial_t a(t) = ad^\bot_{a(t)}a(t)$$
where the adjoint $ad_X^\bot:\mathfrak{sl}_2\to \mathfrak{sl}_2$ is with respect to the inner product $\langle\quad,\quad\rangle_e$ so that
$\langle ad_X^\bot Y,Z\rangle_e = \langle Y,[X,Z]\rangle_e$.

Moreover, for $X\in \mathfrak{so}_2$, the left invariant vector field $L_X$ on $SL_2$ is the infinitesimal generator for the right $SO_2$ action; the momentum mapping (for the lifted action) implies that $\langle X, a(t)=g(t)^{-1}.g'(t)\rangle_e$ is constant in $t$ for each geodesic $g(t)$ on $SL_2$. If the constant is 0 you get horizontal geodesics, and going down to $\mathbb H^2$ is at the same time the Riemannian orbit space and symplectic reduction.
If the constant is not 0, you get a conservation law for ballistic curves.