Python 2 by Wheat Wizard (fourth iteration)

import sys
if set("".join(open(__file__).read().split('\n')[4:]))-set(' &)(,.:[]a`cdfijmonrt~'):sys.setrecursionlimit(1)
for m in sys.modules:sys.modules[m]=None
del sys;f=lambda\
c,d:(`int([]in[])`[:[]in[]]).join([((c)and`dict([((int([]in[])),(int([]in[])))])`[(int([]in[[]])):][(int([]in[[]]))].join([`dict([((int([]in[])),(int([]in[])))])`[(int([]in[]))],`dict([((int([]in[])),c)])`[(int([]in[[]])):][(int([]in[[]])):][(int([]in[[]])):][(int([]in[[]])):]])or`int([]in[])`[:[]in[]]).format((int([]in[]))),((d)and`dict([((int([]in[])),(int([]in[])))])`[(int([]in[[]])):][(int([]in[[]]))].join([`dict([((int([]in[])),(int([]in[])))])`[(int([]in[]))],`dict([((int([]in[])),d)])`[(int([]in[[]])):][(int([]in[[]])):][(int([]in[[]])):][(int([]in[[]])):]])or`int([]in[])`[:[]in[]]).format((int([]in[]))),`(int([]in[]))`]).rfind(`(int([]in[]))`)

Try it online!

No exploits, just a function to add using only characters ' &)(,.:[]a`cdfijmonrt~', as intended (actually only '(),.:[]`acdfijmnort').

I made no attempt to make it short; I just wrote subexpressions for intermediate values like 0 and the empty string and string-substituted those in.

def f(c,d):
	FALSE = []in[]
	TRUE = []in[[]]
	ZERO = int([]in[])
	ONE = int(TRUE)
	EMPTY = `int([]in[])`[:[]in[]]
	ZERO_STR = `ZERO`
	ONE_STR = `ONE`

	ZERO_DICT = dict([(ZERO,ZERO)])
	ZERO_DICT_STR = `ZERO_DICT`

	OPEN_BRACE = ZERO_DICT_STR[ZERO]
	COLON = ZERO_DICT_STR[ONE:][ONE]
	CLOSE_BRACE = ZERO_DICT_STR[ONE:][ONE:][ONE:][ONE:][ONE]

	C_STR = `c`
	D_STR = `d`

	FORMAT_STR_C = ''.join([OPEN_BRACE, ZERO_STR, COLON, C_STR, CLOSE_BRACE])
	FORMAT_STR_D = ''.join([OPEN_BRACE, ZERO_STR, COLON, D_STR, CLOSE_BRACE])

	LENGTH_C_STR = c and FORMAT_STR_C.format(ONE_STR) or EMPTY
	LENGTH_D_STR = d and FORMAT_STR_D.format(ONE_STR) or EMPTY

	TOTAL_STR = EMPTY.join([LENGTH_C_STR, LENGTH_D_STR, ZERO_STR])
	RESULT = TOTAL_STR.find(ZERO_STR)

	return RESULT

Try it online!

The core idea is that string format '{0:5}'.format('1') pads the number zero to a length of 5 like '1 '. By concatenation two such strings using ''.join, the sum of their length is the sum of the input numbers. Then, we tack on a 0 to the end and call .find() to the final position, which is the sum.

The string '{0:5}' to format is produced by extracting the {:} characters from string reprs of dictionaries, created with dict. The string repr of each successive summand is placed where the 5 would be. I wanted to use a dict like {0:5} itself, but its repr includes a space that messed it up.

Inputs of 0 mess up the process because the string sub has a minimum length of 1. We have those with an and/or to give the empty string in this case.