You actually want to check the nested arrays for "1".. you could do that like this

var first_sudoku = [
      ["X", "X", "X", "X"],
      ["1", "X", "2", "X"],
      ["X", "1", "4", "X"],
      ["2", "X", "X", "1"],
];

var found = -1;

for (var i = 0; i < first_sudoku.length; i++) {
    if(first_sudoku[i].indexOf("1")) {
        found = i;
        break;
    }
}

which would set found to the first position in the original array that had the item with "1" in it, or found would be -1 if not found.. so a bit like .indexOf

You can try to add for of loop in your function

var first_sudoku = [
  ["X", "X", "X", "X"],
  ["1", "X", "2", "X"],
  ["X", "1", "4", "X"],
  ["2", "X", "X", "1"],
];

function counter_for_numbers_in_chart(sudoku) {
    for (const row of sudoku) {
        console.log(row.indexOf("1"));   
    }
}

counter_for_numbers_in_chart(first_sudoku);

What you have is an array of arrays. Since "1" doesn't exist in the array you called it on (which only has other arrays in it), you get -1.

In a comment you've said your actual goal is to count all of the "1" elements in the arrays. You'd use a nested loop for that:

let count = 0;
for (const row of first_sudoku) {
    for (const value of row) {
        if (value === "1") {
            ++count;
        }
    }
}

Live Example:

const first_sudoku = [
    ["X", "X", "X", "X"],
    ["1", "X", "2", "X"],
    ["X", "1", "4", "X"],
    ["2", "X", "X", "1"],
];

let count = 0;
for (const row of first_sudoku) {
    for (const value of row) {
        if (value === "1") {
            ++count;
        }
    }
}
console.log(count); // 3