Why is $\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(S^1) \times \rm{Diff}(S^1)$ and not $ \rm{Diff}(\mathbb{R}) \times \rm{Diff}(\mathbb{R})$? The Minkowski metric for $\mathbb{R}^{1,1}$ is
$$
ds^2 = dt^2 - dx^2 = du dv
$$
for coordinates
$$
u = t + x \hspace{1cm} v = t - x
$$
If you do any coordinate transformation that acts independently one $u$ and $v$,
$$
u = f(u') \hspace{1 cm} v = g(v')
$$
then the metric transforms as
$$
ds^2 = du dv = d(f(u')) d(f(v')) = f'(u') g'(v') du' dv'.
$$
Notice that the metric in $(u', v')$ coordinates is just a local rescaling of the original metric $dudv$. Therefore this is a conformal transformation. In fact, all conformal transformations of $\mathbb{R}^{1,1}$ are of this form.
Why then do I often see written 
$\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(S^1) \times \rm{Diff}(S^1)$? Shouldn't it just be $\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(\mathbb{R}) \times \rm{Diff}(\mathbb{R})$? Why is $\mathbb{R}$ "compactified" into $S^1$?
 A: TL;DR: The compactified definition follows a long & useful tradition in e.g. projective & conformal geometry, where one wants to treat infinity on the same footing as a point.
The upshot is that the local conformal groupoid
$${\rm LocConf}(1,1)\tag{A}$$ 
for the 1+1D Minkowski plane is defined in mathematics as the set of locally defined conformal transformations on its conformal compactification$^1$
$$ \overline{\mathbb{R}^{1,1}}~\cong~\mathbb{S}^1\times \mathbb{S}^1 .\tag{B}$$
${\rm LocConf}(1,1)$ contains 4 connected components. The connected component that contains the identity element is
$$ {\rm LocConf}_0(1,1)~\cong~ {\rm LocDiff}^+(\mathbb{S}^1)\times {\rm LocDiff}^+(\mathbb{S}^1).\tag{C}$$
For more details, see e.g. Ref. 1 and this & this Phys.SE posts.
References:


*

*M. Schottenloher, Math Intro to CFT, Lecture Notes in Physics 759, 2008; p. 37.


--
$^1$ It should be mentioned that the conformal compactification contains closed timelike loops, cf. above comment by Cinaed Simson.
