k solution:

{x@<(+/`i$')'x}{x@<x}@" "\:

Explanation:

                      " "\: /split by spaces
               {x@<x}@      /string sort
{   (      )'x}             /for each
       `i$'                 /    cast each char to integer
     +/                     /    sum
 x@<                        /sort by these

Haskell solution (requires the packages sort and extra):

import Data.Sort (sortOn)
import Data.Tuple.Extra ((&&&))

weightSort :: String -> String
weightSort = unwords . sortOn (sum . map (read . pure) &&& id) . words

Python solution:

def sort_by_weight(inputline):
    return " ".join(sorted(inputline.split(), 
                           key=lambda i: (sum(int(j) for j in i), i) ))

And in D, for good measure. I like how easy it is to literally translate Python into D:

import std;

auto sort_by_weight(string inputline) {
  return inputline.split
    .schwartzSort!(x => tuple(x.map!(i => int(i - '0')).sum, x)).join(" ");
}

Btw, “kata” is a Japanese loanword. For the plural, just add s, “katas”.

@raymii pop_back on an empty string is undefined[1] so your solution would probably behave correctly most of the time on empty string as input but might just crash too

[1] https://en.cppreference.com/w/cpp/string/basic_string/pop_back

Fun little exercise. Here’s a Perl solution that works for the test input: https://pastebin.com/yiesqUbg