What is the universal cover of SL(2,R)?

The maximal compact subgroup of $SL(n,\mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,\mathbb{R}) \cong SO(n)\times \mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$\pi_1(SL(n,\mathbb{R})) = \pi_1(SO(n)) =\begin{cases} \mathbb{Z}_2, & n\geq 3; \\ \mathbb{Z}, & n=2.\end{cases}$$
This implies that $\widetilde{SL}(n,\mathbb{R})$ is a two-sheeted cover for $n\geq 3$ and an infinite cyclic cover for $n = 2$.

As for what group it is, well, it's the universal cover of $SL(n,\mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.

For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $\sqrt{2}$. We just haven't met it before so the only way we know it is because $\sqrt{2}\sqrt{2} = 2.$ Same here: $\widetilde{SL}(n,\mathbb{R})$ is the group which is simply connected and covers $SL(n,\mathbb{R})$. That is its defining characteristic.

John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $\tilde{Sp}(2n,{\mathbb R})$ of $Sp(2n,{\mathbb R})$ for any $n \geq 1$.