JavaScript function call inside another call allow you to access arguments that the “arguments[num]” can’t access

i was learning in free code camp and i fell into a problem arguments[1] result into undefined when the function gets called like this functionName(arg1) (arg2) so for example arguments[1] would result in undefined but using this solved it return (second) => addTogether(first, second); and i don’t know how a function call in the function itself could have accessed something that even the arguments[1] couldn’t access, can anyone explain
here is the full code if anyone is interested:

function addTogether() {
  const [first, second] = arguments;
  if (typeof(first) !== "number")
    return undefined;
  if (second === undefined)
    return (second) => addTogether(first, second);
  if (typeof(second) !== "number")
    return undefined;
  return first + second;
}