Ubuntu – How to replace text between two patterns on different lines

First of all, your grep -Po '"\K[^"]*' file idea fails because grep sees both "One" and ". the second is here" as being inside quotes. Personally, I'd probably just do

$ grep -oP '"[^"]+"' file | tr -d '"'
One
Two 
 Three 
Four

But that is two commands. To do it with a single command, you could use one of:

1. Perl

   $ perl -lne '@F=/"\s*([^"]+)\s*"/g; print for @F' file 
   One
   Two 
   Three 
   Four
   

   Here, the @F array holds all matches of the regex (a quote, followed by as many non-" as possible until the next "). The print for @F just means "print each element of @F.

2. Perl

   $ perl -F'"' -lne 'for($i=1;$i<=$#F;$i+=2){print $F[$i]}' file 
   One
   Two 
    Three 
   Four
   

   To remove leading/trailing spaces from each match, use this:

   perl -F'"' -lne 'for($i=1;$i<=$#F;$i+=2){$F[$i]=~s/^\s*|\s$//; print $F[$i]}' file 
   

   Here, Perl is behaving like awk. The -a switch causes it to automatically split input lines into fields on the character given by -F. Since I have given it ", the fields are:

   $ perl -F'"' -lne 'for($i=0;$i<=$#F;$i++){print "Field $i: $F[$i]"}' file 
   Field 0: first matched is 
   Field 1: One
   Field 2: . the second is here
   Field 3: Two 
   Field 0: and here are in second line
   Field 1:  Three 
   Field 2: 
   Field 3: Four
   Field 4: .
   

   Because we are looking for text between two consecutive field separators, we know we want every second field. So, for($i=1;$i<=$#F;$i+=2){print $F[$i]} will print the ones we care about.

3. The same idea but in awk:

   $ awk -F'"' '{for(i=2;i<=NF;i+=2){print $(i)}}' file 
   One
   Two 
    Three 
   Four
   

Updated 18-Nov-2016 (since grep behavior is changed: grep with -P parameter now doesn't support ^ and $ anchors [on Ubuntu 16.04 with kernel v:4.4.0-21-generic])(wrong (non-)fix)

$ grep -Pzo "begin(.|\n)*\nend" file
begin
Some text goes here.  
end

note: for other commands just replace the '^' & '$' anchors with new-line anchor '\n' ______________________________

With grep command:

grep -Pzo "^begin\$(.|\n)*^end$" file

If you want don't include the patterns "begin" and "end" in result, use grep with Lookbehind and Lookahead support.

grep -Pzo "(?<=^begin$\n)(.|\n)*(?=\n^end$)" file

Also you can use \K notify instead of Lookbehind assertion.

grep -Pzo "^begin$\n\K(.|\n)*(?=\n^end$)" file

\K option ignore everything before pattern matching and ignore pattern itself.
\n used for avoid printing empty lines from output.

Or as @AvinashRaj suggests there are simple easy grep as following:

grep -Pzo "(?s)^begin$.*?^end$" file

grep -Pzo "^begin\$[\s\S]*?^end$" file

(?s) tells grep to allow the dot to match newline characters.
[\s\S] matches any character that is either whitespace or non-whitespace.

And their output without including "begin" and "end" is as following:

grep -Pzo "^begin$\n\K[\s\S]*?(?=\n^end$)" file # or grep -Pzo "(?<=^begin$\n)[\s\S]*?(?=\n^end$)"

grep -Pzo "(?s)(?<=^begin$\n).*?(?=\n^end$)" file

see the full test of all commands here (_{out of dated as grep behavior with -P parameter is changed})

Note:

^ point the beginning of a line and $ point the end of a line. these added to the around of "begin" and "end" to matching them if they are alone in a line.
In two commands I escaped $ because it also using for "Command Substitution"($(command)) that allows the output of a command to replace the command name.

From man grep:

-o, --only-matching
      Print only the matched (non-empty) parts of a matching line,
      with each such part on a separate output line.

-P, --perl-regexp
      Interpret PATTERN as a Perl compatible regular expression (PCRE)

-z, --null-data
      Treat the input as a set of lines, each terminated by a zero byte (the ASCII 
      NUL character) instead of a newline. Like the -Z or --null option, this option 
      can be used with commands like sort -z to process arbitrary file names.