The Earth's curvature IS 8 inches per mile BUT that does not mean the visible horizon has any direct correlation with that value (i.e. the "drop" in the horizon over 10 miles is not 80 inches). I was wrong before in simplifying it to that. The real way to determine the horizon "drop" is Pythagorean Theorem (remember that?).
Anyway, you're correct that over 10 miles there is a "hump" of 50-60 feet, but the answer to why you can see something at or beyond that distance is simple refraction. Light bends in different mediums and the atmosphere is a smorgasbord of disturbances and density changes. Physics (and science generally) is a wonderful thing and I sometimes feel like many of these flat-earthers are too obtuse to appreciate it.
For further reference this table gives general corrections for Earth's curvature and refraction.
View attachment 36484
Source:
http://armyengineer.tpub.com/en0593a/en0593a0085.htm
Forgive me it's metric (you Yanks really need to get on board), but as you can see at a distance of 16km (10 miles) the refraction effect is 17.3 metres (~57 feet). This eliminates the "hump" caused by the Earth's curvature over Lake Champlain.