Roughly because the number of times it needs to sum to get a multiply is not constant; it depends on the input. Think about writing the automaton -- how would you know when to stop? You're only given one automaton to write, and you don't know ahead of time how big either of the input numbers are.
So how do you get it to add? Specifically, what representation of arbitrary numbers are we using with our finite automaton? Infinite read-only tape plus infinite write-only tape for output, in base 10? If so, it should be simple to do multiplication by representing the numbers in factorized form and applying our adder.