Z80 DAA instruction

Does it just modify the accumulator anyway, based on the conditions set out in the DAA table, regardless of the previous instruction?

Yes. The documentation is only telling you what DAA is intended to be used for. Perhaps you are referring to the table at this link:

--------------------------------------------------------------------------------
|           | C Flag  | HEX value in | H Flag | HEX value in | Number  | C flag|
| Operation | Before  | upper digit  | Before | lower digit  | added   | After |
|           | DAA     | (bit 7-4)    | DAA    | (bit 3-0)    | to byte | DAA   |
|------------------------------------------------------------------------------|
|           |    0    |     0-9      |   0    |     0-9      |   00    |   0   |
|   ADD     |    0    |     0-8      |   0    |     A-F      |   06    |   0   |
|           |    0    |     0-9      |   1    |     0-3      |   06    |   0   |
|   ADC     |    0    |     A-F      |   0    |     0-9      |   60    |   1   |
|           |    0    |     9-F      |   0    |     A-F      |   66    |   1   |
|   INC     |    0    |     A-F      |   1    |     0-3      |   66    |   1   |
|           |    1    |     0-2      |   0    |     0-9      |   60    |   1   |
|           |    1    |     0-2      |   0    |     A-F      |   66    |   1   |
|           |    1    |     0-3      |   1    |     0-3      |   66    |   1   |
|------------------------------------------------------------------------------|
|   SUB     |    0    |     0-9      |   0    |     0-9      |   00    |   0   |
|   SBC     |    0    |     0-8      |   1    |     6-F      |   FA    |   0   |
|   DEC     |    1    |     7-F      |   0    |     0-9      |   A0    |   1   |
|   NEG     |    1    |     6-F      |   1    |     6-F      |   9A    |   1   |
|------------------------------------------------------------------------------|

I must say, I've never seen a dafter instruction spec. If you examine the table carefully, you will see that the effect of the instruction depends only on the C and H flags and the value in the accumulator -- it doesn't depend on the previous instruction at all. Also, it doesn't divulge what happens if, for example, C=0, H=1, and the lower digit in the accumulator is 4 or 5. So you will have to execute a NOP in such cases, or generate an error message, or something.

Just wanted to add that the N flag is what they mean when they talk about the previous operation. Additions set N = 0, subtractions set N = 1. Thus the contents of the A register and the C, H and N flags determine the result.

The instruction is intended to support BCD arithmetic but has other uses. Consider this code:

    and  15
    add  a,90h
    daa
    adc  a,40h
    daa

It ends converting the lower 4 bits of A register into the ASCII values '0', '1', ... '9', 'A', 'B', ..., 'F'. In other words, a binary to hexadecimal converter.