C# solution with comments:

public static string counterGame(long n)

    {

            bool currentPlayer1 = true;


    while(n > 1)
    {   
        // Check if n is a power of 2
        //A number is a power of 2 if it has exactly one 1 bit in its binary representation. (i.e. 100)         
        if((n & (n - 1)) == 0)
        {
            n = n/2;             
        }
        else
        {             
            //Math.Log(n, 2) -  Math.Log(20, 2) will return approximately 4.32 because 2 4.32 = 20
            //Math.Floor(4.32) round down, so will produce 4.
            //Math.Pow(2, 4) which will produce 16
            //So finally 20 - 16, will give 4, which will be the new number for the next turn.
            long largestPowerOf2 = (long)Math.Pow(2, (long)Math.Floor(Math.Log(n, 2)));
            n -= largestPowerOf2;                                
        }

        currentPlayer1 = !currentPlayer1;

    }

    return currentPlayer1 ? "Richard" : "Louise";
}

Javascript

function counterGame(n) {
    // Write your code here    
    var turn = 0
    while(n > 1){
        n = Math.log2(n) % 1 == 0 ? n/2 : n - Math.pow(2, Math.floor(Math.log2(n)))        
        turn ++
    }
    return turn % 2 == 0 ? 'Richard' : 'Louise'
}

Java solution class Result {

public static String counterGame(long n) {
    if (n == 1) return "Richard";

    int turn = 0;
    while (n > 1) {
        if ((n & (n - 1)) == 0) { // Check if n is a power of two
            n /= 2;
        } else {
            n -= Long.highestOneBit(n); // Get the largest power of 2 less than or equal to n
        }
        turn++;
    }
    return (turn % 2 == 0) ? "Richard" : "Louise";
}

}

Javascript

function counterGame(n) {
    let turn = 0;
    let log = Math.log2(n);
    while (log % 1 !== 0) {
        n -= Math.pow(2, Math.floor(log)); //subtract nearest power of 2
        turn++;
        log = Math.log2(n);
    }
    const endTurn = turn + log;
    return endTurn % 2 === 0 ? "Richard" : "Louise";
}