If you're maintaining the same speed in both cases - meaning you stay at a steady speed all the way up - it takes more power to climb the hill faster. Therefore, the engine will be producing more power and applying more power to the input of the transmission the faster you go.
As far as torque applied to various parts of the transmission - that would take a lot of calculations and you'd have to do that to try to figure out which one is better/worse. For example if the first scenario required 40hp, but was turning a shaft internally at 2000rpm, then the torque on that shaft would be 105 ft-lbs. If the second scenario requires 80hp, but the internal shaft is spinning at 7000rpms, then that shaft will experience a torque of only 60 ft lbs. So it would have less torque applied while higher horse power.
So - the point there is that it's possible to have a transmission apply more power with less force on an internal part. BUT that's pretty rare in most cases. Normally most parts rotate slower than or equal to the input shaft speed (except in overdrive, but that's not really meaningful here). But I think the general rule in most cases the more power transferred is normally going to translate to more heat, more force on the input parts, and more wear. The least wear is probably to speed up on the flat ground slowly, and then use the momentum and drive up as the vehicle slows down.
If your elevation gain is 400 feet in 1/4 mile - ballpark numbers, if you're going to climb the hill at 30mph, it'll require an extra 120hp over flat ground. If you're going to go 10mph, it'll take an extra 40hp. I think you could assume that it's likely that pushing another 120hp through the transmission will probably cause more wear than 40hp. That's not a certainty, but I think in this case without knowing a lot more about the transmission internals to tell me something that would suggest otherwise from what I would normally guess to be the case, I'd say going slower is easier on the transmission than going fast.