awk: ^ backslash not last character on line

Rather than testing over ssh, you can replicate the behaviour using eval. I made a test file (called month):

Mar line1 line2 line2 Apr line3 

You have (at least) three options:

First option

Your two options are mutually exclusive, so you can sidestep the issue of escaping a ! entirely by using two blocks with next in the first block:

eval "awk '/^Mar|^Apr/ { printf(\"\\n%s\",\$0); next } { printf(\"%s\", \$0) }' month" 

If the condition is true, the first block is taken and next skips the rest. Note that I have removed the unnecessary $0 ~ from the condition. The match is performed against the whole line by default.

Second option

You could actually just do this:

eval "awk '/^Mar|^Apr/ { \$0 = \"\\n\"\$0 } { printf(\"%s\", \$0) }' month" 

1. If the line matches, precede it with a newline.
2. In all cases (no condition before the { }), print the line.

Third option

If you wrap the overall command in single quotes, you don't need to do anything fancy with the !:

eval 'awk "{ if(/^Mar|^Apr/) { printf(\"\\n%s\",\$0) } if(!/^Mar|^Apr/) { printf(\"%s\", \$0)} }" month' 

I recommend one of the other two solutions, I just thought that it would be worth showing that you can use ! within the command if you need to.

Output for all three cases:

Mar line1line2 line2 Apr line3 

Here is a cleaned up version of your awk

awk '/^Mar|^Apr/ {printf "\n%s",$0;next} {printf "%s",$0}' 

This will test if $0 (default, so need to add) is starting with Mar or Apr
If yes do printf "\n%s",$0;next.
The next makes final code to be rune only if $0 is not starting with Mar or Apr
Then run printf "%s",$0

Or just:

awk '/^Mar|^Apr/ {$0="\n"$0} {printf "%s",$0}'