Stripping quotes in a for / findstr loop

I am trying to process a text file that looks like this:

"N" "accounts" "General" no "0001"
"N" "accounts" "Customer" no "0002" 
and so on

The batch file looks like this:

setlocal EnableDelayedExpansion 

set TGT=schctrl
set TMP=schctrl.tmp

copy %TGT% %TMP%

set NWOV=accounts inventory generalledger headoffice

for %%a in (%NWOV%) do (
   
   for /f "tokens=2,5 delims= " %%D in ('findstr /i %%a %TMP%') do (
      set M=%%E
      set M=%M:"=%
      echo %%D %%E %M%
   )
)

The command, set M=%M:"=%, doesn't strip off all the quotation marks!

When I run it, I get output like this:

set M="1148"
set M=!M:"=!
echo "accounts" "1148"

I am expecting the last line to look like this:

"accounts" "1148" 1148

The last field is missing.

I am guessing the culprit is the for loop, and expansion of variables, but I have been grinding away on this and have not yet found the solution.

When you open cmd and run for /?. Scroll down to the section on Substitution and read the first line: Therefore %%E will contain quotes and %%~E won't contain quotes. So, just changing that meta variable in the code:

@echo off

set "TGT=schctrl"
set "TMP_FILE=schctrl.tmp"

copy "%TGT%" "%TMP_FILE%"

set "NWOV=accounts inventory generalledger headoffice"

for %%a in (%NWOV%) do (
   for /f "tokens=2,5 delims= " %%D in ('findstr /i %%a %TMP_FILE%') do (
      echo %%D %%E %%~E
   )
)

As a side note, the reason why your search/replace did not work is because you enabled delayedexpansion but you never used it. For that you needed to replace % with ! in the variables you are expanding within the loop, so !M! instead of %M%

That being said, please do not use single characters as variable names it will create confusion. Also, as stated by Magoo in the comments, do not re-use system variables, like %TMP%.