Much of this has been covered in other Q&As (especially the later C++ How is release-and-acquire achieved on x86 only using MOV?), but I'll give a summary here. Still, good question, it's useful to collect this all in one place.
On x86, every asm load is an acquire-load. To implement that efficiently, modern x86 HW speculatively loads earlier than allowed and then checks that speculation. (Potentially resulting in a memory-order mis-speculation pipeline nuke.) To track this, Intel calls the combination of load and store buffers the "Memory Order Buffer".
Weakly-ordered ISAs don't have to speculate, they can just load in any order.
x86 store ordering is maintained by only letting stores commit from the store buffer to L1d in program order.
On Intel CPUs at least, a store-buffer entry is allocated for a store when it issues (from the front-end into the ROB + RS). All uops need to have a ROB entry allocated for them, but some uops also need to have other resources allocated, like load or store buffer entries, RAT entries for registers they read/write, and so on.
So I think the store buffer itself is ordered. When a store-address or store-data uop executes, it merely writes an address or data into its already-allocated store-buffer entry. Since commit (freeing SB entries) and allocate are both in program order, I assume it's physically a circular buffer with a head and tail, like the ROB. (And unlike the RS).
Avoiding LoadStore is basically free: a load can't retire until it's executed (taken data from the cache). A store can't commit until after it retires. In-order retirement automatically means that all previous loads are done before a store is "graduated" and ready for commit.
A weakly-ordered uarch that can in practice do load-store reordering might scoreboard loads as well as tracking them in the ROB: let them retire once they're known to be non-faulting but, even if the data hasn't arrived.
This seems more likely on an in-order core, but IDK. So you could have a load that's retired but the register destination will still stall if anything tries to read it before the data actually arrives. We know that in-order cores do in practice work this way, not requiring loads to complete before later instructions can execute. (That's why software-pipelining using lots of registers is so valuable on such cores, e.g. to implement a memcpy. Reading a load result right away on an in-order core destroys memory parallelism.)
How is load->store reordering possible with in-order commit? goes into this more deeply, for in-order vs. out-of-order.
Barrier instructions
The only barrier instruction that does anything for regular stores is mfence which in practice stalls memory ops (or the whole pipeline) until the store buffer is drained. Are loads and stores the only instructions that gets reordered? covers the Skylake-with-updated-microcode behaviour of acting like lfence as well.
lfence mostly exists for the microarchitectural effect of blocking later instructions from even issuing until all previous instructions have left the out-of-order back-end (retired). The use-cases for lfence fo memory ordering are nearly non-existent.
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