Here's one way to do this. We divide by 10, several times in a row if necessary, store the remainders, and then adjust them and output them in the reverse order.
We start with 0 0 0 x 0 0 0 0 0 (we may need up to 9 cells). Assume we start with the pointer at the leftmost 0 here.
While doing the first division, the memory layout will be: 0 remainder divisor dividend quotient 0 0 0 0.
Then if the quotient is nonzero, we use it as the next dividend, so during the second division the layout will be "0 1st_remainder 2nd_remainder divisor dividend quotient 0 0 0".
We will increase remainders by 1 so we can easily tell them apart from empty space, to know how long the result is.
Also: to print 0 properly as "0" we'll set it up as if it were a remainder, and then if x is nonzero we'll clear it out again and compute remainders properly.
About how the division works: the basic principle here is that we have an outer loop executed (dividend) times, and on the nth time through it sets remainder and quotient to the proper values for (n/divisor). Usually this just means incrementing the remainder, and decrementing the divisor because we're using it as a counter. Each tenth time through, we need to zero the remainder and increment the quotient; we use the remainder to refill the divisor counter at the same time; we do this whenever the divisor counter has been zeroed.
>+>>[ set up fake remainder for 0
<<->>[ x is nonzero; clear fake remainder and start main loop
<++++++++++> set divisor to 10
[<<+>-[>>>>]<[[>+<-]>>>+>]<<-] divide
<<+>[-]>> increment remainder and clear divisor
]<
]<<[
->++++++++[<++++++>-] increase by 47
<.[-]< output digit and clear
]++++++++++.[-] output linefeed and clear
