Not really, only if you are betting twice as much. If you're average bet would be normally 10 dollars, and you instead bet 5 dollars on each of those, it's equivalent to one 10 dollar bet.

This applies more generally to all bets in roulette. Every single bet in roulette has the exact same house edge and can often be replicated by multiple bets of a different one. I.E. one double number bet has the same odds as two single number bets of the same amount. Making two dozen number bets is equivalent to a hypothetical 24 number bet that just isn't normally offered at the table.

Basically, how much you lose at roulette is 100% based on the volume of your overall bet. Which numbers/bets you select does not matter. The only thing it matters towards is variance. Some people want to have rare large wins and frequent small losses, some people want frequent small wins and rare large losses. The selection of numbers you pick will affect that, but again not the expected amount lost.

That's only because in your example he's betting twice as much. A two unit bet on black is the same house edge as two one unit bets on different dozen bets.

It's just an adjustment on how often you win and how much each win is. There is a wide variety in the casino on how much different bets pay out and how likely they are to win. There are bets that are very unlikely to win and pay out a lot when they do, and there are bets that pay very little but are likely to win. This particular set of bets leans heavily to the latter, it only pays 1 to 2, but wins almost 2/3s of the time. It's overall house edge is the same as any other bet on the roulette table.

It is fair to say this bet has lower variance compared to other bets, which is what most people look for in gambling, but if this guy enjoys betting that way, there is nothing wrong with it. He won't lose any more in the long run than a person betting black every spin.