How to group the following strings by regular expression

You could exclude dot and underscore from matching.

var string = '_A._B._C._D._F.f',
    result = string.match(/[^._]+/g);

console.log(result);

How about that without using regex?

str = '_A._B._C._D._F.f'.split('.')
var alphabets = str.map(c => c.replace('_', ''));
console.log(alphabets);

You can use split that removes [._]+ (any substring containing dots or floors) and the filter (to remove the initial empty string):

'_A._B._C._D._F.f'.split(/[._]+/).filter(function(s){ return s.length > 0})
# => [ "A", "B", "C", "D", "F", "f" ]

EDIT: Simplification suggested in comments:

'_A._B._C._D._F.f'.split(/[._]+/).filter(Boolean)
# =>  [ "A", "B", "C", "D", "F", "f" ]

In your regex you try to match the whole pattern using an anchor ^ to assert the start of the string followed by a character class which will match only one out of several characters (and might for example also be written as [_(+\\.)]+) and then you capture the rest of the string in a capturing group and assert the end of the line $.

If you want to check the format of the string first, you might use a more exact pattern. When that pattern matches, you could do a case insensitive match for a single character as the pattern is already validated:

const regex = /^_[A-Z](?:\._[A-Z])+\.[a-z]$/;
const str = `_A._B._C._D._F.f`;

if (regex.test(str)) {
  console.log(str.match(/[a-z]/ig));
}

See the regex demo

That will match:

• ^ Assert the start of the strin
• _[A-Z] Match an underscore and an uppercase character
• (?:\._[A-Z])+ 1+ times repeated grouping structure to match ._ followed by an uppercase character
• \.[a-z] Match a dot and a lowercase character
• $ Assert the end of the line

string method .match with global flag, can help you:

console.log('_A._B._C._D._F.f'.match(/[a-z]+/gi))