address field and words of memory
Let's start with the word "word" (pun intended)

In this case it represents the default size of the storage medium of the system.  This could be any number of bits, but was commonly 8 bits (in, for example the Z80 of the ZX Spectrum, etc), or 16 bits in the early PC systems (8086, 80286 etc).

So an 8 bit computer has a word size of 8 bits.

Then there are 16 address bits.  This is literally the number of address lines on the chip.  Again, taking the Z80 as an example, there were 16 of them (A0 to A15).  This gives a possible \$2^{16}\$ addresses - 65536.

Each one of those addresses represents the memory location of one word of data.

That's 65535 available words - on an 8-bit system that's 64KBytes.  On a 16-bit system it would be double that at 128KBytes.

Now, the RAM memory, the ROM memory, and (depending on the architecture) the IO peripherals will all take a number of those 65536 addresses.  

Say for example you have 2K of ROM and 4K of RAM.  That's 2048 addresses of ROM and 4096 addresses of RAM.  Not all the addresses are used up, so there is room for memory expansion say.

The ZX Spectrum 48K had 16K of ROM and 48K of RAM for example.  That's 16384 addresses pointing to ROM, and 49152 addresses pointing to RAM.

How much of the available 65536 addresses are actually used is purely down to the designer of the computer.