What do the brackets mean in x86 asm?

[L1] means the memory contents at address L1. After running mov al, [L1] here, The al register will receive the byte at address L1 (the letter 'w').

Operands of this type, such as [ebp], are called memory operands.

All the answers here are good, but I see that none tells about the caveat in following this as a rigid rule - if brackets, then dereference, except when it's the lea instruction.

lea is an exception to the above rule. Say we've

mov eax, [ebp - 4]

The value of ebp is subtracted by 4 and the brackets indicate that the resulting value is taken as an address and the value residing at that address is stored in eax. However, in lea's case, the brackets wouldn't mean that:

lea eax, [ebp - 4]

The value of ebp is subtracted by 4 and the resulting value is stored in eax. This instruction would just calculate the address and store the calculated value in the destination register. See What is the difference between MOV and LEA? for further details.

Simply means to get the memory at the address marked by the label L1.

If you like C, then think of it like this: [L1] is the same as *L1

The brackets mean to de-reference an address. For example

mov eax, [1234]

means, mov the contents of address 1234 to EAX. So:

1234 00001

EAX will contain 00001.

Direct memory addressing - al will be loaded with the value located at memory address L1.