Sed string substitution with Shell Variable

The $# is a variable in shell with number of arguments.

$ ( set -x ; sed -r "s#(.*/)([[:digit:]]+)([[:space:]]+/)$#\112\3#" )
+ sed -r 's#(.*/)([[:digit:]]+)([[:space:]]+/)0\112\3#'
sed: -e expression #1, char 44: unterminated `s' command

So $# is replaced by 0 (in case of my shell, it's interactive, has no arguments).

Research difference between single and double quotes. Research when to quote in shell. Research how quotation works in shell. You seem to want:

sed -r 's#(.*/)([[:digit:]]+)([[:space:]]+/)$#\1'"$REPOVER"'\3#'

Based on the sample data provided the proposed sed can be simplified a bit:

$ sed -r "s|/[0-9]+ |/${REPOVER} |" testing.list
$ sed -r "s|/[[:digit:]]+ |/${REPOVER} |" testing.list

Find and replace the pattern /<at_least_one_number><space> with /${REPOVER}<space>

Both of which produce:

deb [trusted=yes] http://10.47.4.220/repos/test-repo/12  /

Your attempt 2 is almost correct. Just escape the $ signs that you want to preserve from shell expansion:

sed -r "s#(.*/)([[:digit:]]+)([[:space:]]+/)\$#\1$REPOVER\3#" testing.list

Another option is to use a mixture of simple quotes and double-quotes:

sed -r 's#(.*/)([[:digit:]]+)([[:space:]]+/)$#\1'"$REPOVER"'\3#' testing.list

But all this works only if the value of your variable does not contain special characters that sed would interpret. If you have such characters you will need to first escape them. If, for instance, the value of REPOVER can contain #, & or \ characters you can:

ESCAPED_REPOVER="$(echo "$REPOVER" | sed -r 's/(#|\\|&)/\\\1/g')"
sed -r 's#(.*/)([[:digit:]]+)([[:space:]]+/)$#\1'"$ESCAPED_REPOVER"'\3#' testing.list

A last, complicated one, in case REPOVER can really be anything, including endline characters, bash and sed special characters, etc. But let's however assume it does not itself contain lines beginning with deb (if it did we would need a custom delimiter but the principle would be almost the same).

We first print the variable, next cat the file, and pass all this to sed. The sed script first collects the preamble in the hold space, that is, everything until the first ^deb line excluded. It then uses the content of the hold space for the replacement in all ^deb lines it encounters. Non ^deb lines after the preamble are printed unmodified:

{ printf '%s\n' "$REPOVER" ; cat testing.list; } | \
  sed -rn '1,/^deb/{/^deb/!{H;b;}};/^deb/{G;s#[[:digit:]]+([[:space:]]+/)\n\n(.*)#\2\1#;};p'

Demo:

$ cat testing.list 
deb [trusted=yes] http://10.47.4.220/repos/test-repo/11  /
deb [trusted=yes] http://10.47.4.220/repos/foo-repo/11  /
# comment

deb [trusted=yes] http://10.47.4.220/repos/bar-repo/11  /

# another comment
$ printf -v REPOVER '\\$&#'
$ { printf '%s\n' "$REPOVER" ; cat testing.list; } | \
sed -rn '1,/^deb/{/^deb/ba;H;b;};:a;/^deb/{G;s#[[:digit:]]+([[:space:]]+/)\n\n(.*)#\2\1#;};p'
deb [trusted=yes] http://10.47.4.220/repos/test-repo/\$&#  /
deb [trusted=yes] http://10.47.4.220/repos/foo-repo/\$&#  /
# comment

deb [trusted=yes] http://10.47.4.220/repos/bar-repo/\$&#  /

# another comment

Changing the delimiter and placing it in double quotes worked for me.

sed -r "s|(.*/)([[:digit:]]+)([[:space:]]+/)$|\1$REPOVER\3|"