Your numbers above:
Up the river T = 10 nm / (10 kn - 2 kn) = 1.25 hrs = 75 min
Down the river T = 10 nm / (10 kn + 2 kn) = 0.833 hrs = 50 min
Wadden explained how to do the calculations, but Baker asked the more important question of "why?"
RT asks "where did the other 5 minutes go?" which wadden then explained in that the 2 kt difference is spread over different distances in the two examples, which is true.
I look at it another way. In the example of being set by 2 kt current, the effect on your travel time is going to be equal to the percentage of that current to your final SOG. In this case 25%. 2/8=25%. Add 25% to what your travel time would be if you weren't fighting the current, or 15 minutes in this case.
In the case where you are being carried by a current, it is the same thing. The effect on your travel time will be the percentage of the current compared to your SOG. I this case 16.6%. 2/12= 16.6% You reduce your travel time by 16.6% over what it would be without a current. 16.6% of an hour is 10 minutes.
Conceptually, think of it this way, a contrary current is a bigger percentage of your SOG than a helpful current is.
Not to be too pedantic, but... yeah I am being too pendantic.
Use the same example only a 4 knot current.
Boat speed = 10kt
Current = 4kt
Distance = 10nm
Affect on time to destination is current/SOG
Against the current: 4/6= 67%, 67% of 60 minutes is 40 minutes longer to make the trip.
With the current: 4/14= 28%, 28% of 60 minutes is about 17 minutes less time to make the trip.
Bottom line is that current hurts you much more than it helps you if you are looking at not only time, but fuel consumption.
Learning to sail in a region with strong currents on boats with unreliable motors makes you very aware of the effect of currents. It also makes you aware of where the current is. If you are in a boat with a top speed of 5 knots, a 4-6 knot current is a big deal!