(For some reason first I thought that the question was only in dimension $2$...)
The essence of this statement (in dimension $2$) consists of two facts that are useful anyway:
- The index $1$ cover of a klt singularity is canonical (in any dimension): it is relatively easy to see that it is still klt by looking at discrepancies and then by definition of the index $1$ cover the discrepancies are integers, so it is canonical.
- An index $1$ canonical singularity is the same as a rational Gorenstein singularity. In dimension $2$ those are the rational double points or Du Val singularities which have been known to be quotient singularities "classically". (See for instance Durfee, Alan H., Fifteen characterizations of rational double points and simple critical points, L'Enseignement Mathématique. Revue Internationale. IIe Série 25 (1) (1979), 131–163, ISSN 0013-8584 , MR 543555)
For the general case see Prop. 9.3 in Greb, Kebekus, Kovács, Peternell: Differential forms on log canonical spaces, Publ. Math. Inst. Hautes Études Sci. No. 114 (2011), 87–169.
A heuristic way of seeing why this should be true is the following: Take a klt singularity and localize at a codimension $2$ point. You get a klt surface which is a quotient singularity by the above argument. The problem with this approach is that the residue field at the codimension $2$ point is not algebraically closed so you would either have to pass to the algebraic closure or analyze singularities over non-algebraically closed fields and then once you have the localized quotient singularity description prove that it comes from one in the original situation. All of this is possible to do, but not entirely self-evident. The proof referenced above proceeds in a different way.
