Many-One Reducibility of decision problems for complexity theory?
NP-hardness (and so NP-completeness) are defined using many-one reductions. When you show that a problem is NP-hard, you have to follow the definition and use many-one reductions. All proofs that you see of NP-hardness use many-one reductions.

One thing to note is that usually, one does not show how to reduce an arbitrary problem in NP to the target problem. Rather, one reduces a problem already known to be NP-hard to the target problem. The starting point of this endeavor is Cook's theorem (independently proved by Levin), which shows that SAT is NP-hard.

The other common type of reduction is oracle reduction (there are other types as well, but they are less common). Oracle reductions cannot distinguish NP from coNP: every coNP-hard problem in the many-one sense is NP-hard for oracle reductions. Most researchers believe that NP is different from coNP, and this is one of the reasons we insist on using many-one reductions.