How can segment be 60KB
In "real mode" on the x86, an address is 32 bits long. The higher 16 bits can be considered the segment address, while the lower 16 bits are the address within the segment. But it's a bit more complicated than having $2^{16}$ segments of $2^{16}$ addresses each.

If you have a 16-bit address, then you're able to address $2^{16} = 65,536$ different locations. If each of those is a byte, you have 65,536 bytes, or 64KB, of addressable memory. So a segment is at most 64KB long.

To get the actual address, the segment address is effectively shifted left by four bits, then added to the address within the segment. This does, as you've noted, mean that the addresses are not unique! The address 0x0000.00FF is the same as the address 0x000F.000F in this system. Segments can, and frequently do, overlap. And this isn't considered a problem, because in "real mode", there's no protection whatsoever. Segments are just a convenience; the CPU doesn't care if you want to write to opposite ends of memory in completely different segments.

This also limits the total amount of addressable memory to $2^{20}$ bytes, rather than the $2^{32}$ you would expect with an address of that length. But this is also by design: the 8086 has only 20 address pins. So despite the apparent clumsiness of this system, it ended up working out well.