The memory map in the datasheet shows the 126kB occupies 0x80000 to 0x9F800. Which section is Block 0 and which is Block 1 with respect to the status registers FEE0STA and FEE1STA and the control registers FEE0CON0 and FEE1CON0?
As written on page 29 of the data sheet the Flash is 2 blocks of 64kByte.
Only 126kByte are available to the user.
So the first block physical address range is from
0x80000 to 0x8FFFF
and the second user block is from
0x90000 to 0x97FFF
Here is my interpretation.
Combining what you said with the paragraph on page 29, which states that "In the first block, 31 kB x 16 bits is user space and 1 kB x 16 bits is reserved for factory-configured boot page", 0x90000 - 0x9F800 is "BLOCK 0" and FEE0STA applies to this block.
yes, you are right - the text is a bit misleading.
But it is easy to figure out with a simple test - writing via FEE0-MMRs a 0x1234 pattern at a address and another pattern like 0x5678 via FEE1MMRs and looking via debugger into the Flash shows it clearly.
Retrieving data ...