Regarding obtaining a series expansion at infinity
You can use the new in M12 function AsymptoticSolve:
AsymptoticSolve[y == x Erf[x]^2/Erf[2x], y, {x, Infinity, 3}]
{{y -> (E^-x^2 (1 - 2 x^2 + E^x^2 Sqrt[π] x^3))/(Sqrt[π] x^2)}}
You can use the new in M12 function AsymptoticSolve:
AsymptoticSolve[y == x Erf[x]^2/Erf[2x], y, {x, Infinity, 3}]
{{y -> (E^-x^2 (1 - 2 x^2 + E^x^2 Sqrt[π] x^3))/(Sqrt[π] x^2)}}