A \def syntax problem

\numexpr expressions have to expand to the numeric syntax so you can not have unexpandable assignments from \def

I think this is the calculation you intended

\documentclass{article}

\def\nok#1:#2\relax#3:#4\relax{%
\numexpr
 \numexpr(#3)*60+#4\relax
 \ifnum\numexpr(#1)*60+#2\relax<\numexpr(#1)*60+#2\relax+1440\fi
\relax
 }

\begin{document}

\the\numexpr(\nok22:44\relax22:50\relax)\relax

\end{document}

The expected expl3 answer providing the expected padding. This also support (single) macros in the input.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn

\NewExpandableDocumentCommand{\timedelta}{mm}
 {
  \joseph_timediff:ff { #1 } { #2 }
 }

\cs_new:Nn \joseph_timediff:nn
 {
  \__joseph_timediff:ff
   { \__joseph_minutes:w #1 \q_stop }
   { \__joseph_minutes:w #2 \q_stop }
 }
\cs_generate_variant:Nn \joseph_timediff:nn { ff }

\cs_new:Nn \__joseph_timediff:nn
 {
  \joseph_output:f
   {
    \int_eval:n { (#2) - (#1) \int_compare:nNnT { #2 } < { #1 } { + 1440 } }
   }
 }
\cs_generate_variant:Nn \__joseph_timediff:nn { ff }

\use:x
 {
  \cs_new:Npn \exp_not:N \__joseph_minutes:w ##1 \token_to_str:N : ##2 \exp_not:N \q_stop
 }
 {
  #1 * 60 + #2
 }

\cs_new:Nn \joseph_output:n
 {
  \int_compare:nNnT { \int_div_truncate:nn { #1 } { 60 } } < { 10 } { 0 }
  \int_div_truncate:nn { #1 } { 60 }
  :
  \int_compare:nNnT { \int_mod:nn { #1 } { 60 } } < { 10 } { 0 }
  \int_mod:nn { #1 } { 60 }
 }
\cs_generate_variant:Nn \joseph_output:n { f }

\ExplSyntaxOff

\newcommand\routeStart{01:27}
\newcommand\routeStop{01:27}

\begin{document}

\timedelta{22:40}{22:50}

\timedelta{23:00}{00:10}

\timedelta{08:24}{19:32}

\timedelta{\routeStart}{01:27}

\timedelta{16:03}{\routeStop}

\timedelta{\routeStart}{\routeStop}

\end{document}

You can also use Lua, if you're willing to compile with lualatex instead of pdflatex:

\documentclass{article}

\directlua{dofile('timediff.lua')}
\def\wait#1#2{\directlua{tex.print(timediff('#1', '#2'))}}

\begin{document}
The time from 22:44 to 22:50 is \wait{22:44}{22:50}.

The time from 23:00 to 00:10 is \wait{23:00}{00:10}.
\end{document}

where timediff.lua is:

function hhms(s)
   -- Example: hhms('01:42') gives 102
   local _, _, hh, mm = string.find(s, "(%d*):(%d*)")
   return hh * 60 + mm
end

function timediff(time1, time2)
   -- Example: timediff('22:45', '01:15') gives '02:30'
   local t1 = hhms(time1)
   local t2 = hhms(time2)
   if t2 < t1 then t2 = t2 + 1440 end
   local d = t2 - t1
   local mm = d % 60
   local hh = (d - mm) / 60
   return string.format('%02d:%02d', hh, mm)
end

Later you may find Lua's time and date functions useful.