How can I get words between the first two instance of text/pattern?

Here is one in awk:

$ awk '/^=+$/{f=!f;if(f==1)next;else if(f==0)exit}f' file
v2.0.0

Added feature 3
Added feature 4

In pretty print:

$ awk '/^=+$/ {     # at ===...
    f=!f            # flag state is flipped
    if(f==1)        # if its one (first ===...)
        next        # next record
    else if(f==0)   # if zero (second ===...)
        exit        # nothing more to do yeah
}
f' file             # print

Here is another one in GNU sed:

$ sed -n '/^=\+$/,//{//!p;b};q' file
v2.0.0

Added feature 3
Added feature 4

• /^=\+$/,// is a shorthand for /^=\+$/,/^=\+$/, it selects the lines between two lines consisting of equal signs inclusively, and the commands between following curly braces are executed for these lines,
• //!p is a shorthand for /^=\+$/!p, it means if incoming line is not one of those which consist of only =s, print it,
• b means go to the end of cycle (i.e pass q),
• q is for exitting sed after printing selected lines.

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The following version will work with all POSIX-compliant seds but it looks 2x more cryptic:

sed -n -e '/^=\{1,\}$/,//{//!p;b' -e '}' -e 'q' file

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Note that these are not gonna work if there are two consequent all = lines in the input.

Could you please try following too.

awk '/^=/{count++;next} count>=2{exit} {print}'  Input_file