From logical view it looks fine. You could also try the ternary operator if you want to play around.

return n <= 0 ? 0 : sum(arr, n - 1) + arr[n - 1];

The first block is the if question. If its true is goes to the second block (starts whith ?) and if its false it goes to the third block (starts with :).

Actually, with n equals to 1 is quite easy. Let's check step by step. Here your sum function; let's add some line to make it easier to refer:

1: function sum(arr, n) {
2:  if(n <= 0){
3:    return 0;
4:  }else {
5:    return sum(arr, n - 1) + arr[n - 1];
6:  }
7: }

Now, let's see what's happening step by step when you execute:

sum([2, 3, 4], 1)

The function sum is called with arr equals to [2, 3, 4], and n equals to 1.

Since n is not less or equals than 0 (line 2), we go to the else block, at line 5.

Now, here is where the recursion happens: we call again the function sum, passing the same arr, but not the same n, instead we pass n - 1.

So we call sum again, this time with arr equals to [2, 3, 4] but with n equals to 0. Since n is 0 this time, at the line 2 check we proceed to line 3 and returns 0.

Now, the function exit, with the value 0, that we gave to the caller.

And the caller of sum([2, 3, 4], 0) was the execution sum([2, 3, 4], 1), specifically at line 5:

5:    return sum(arr, n - 1) + arr[n - 1];

Since it returned 0, we can imaging like:

5:    return 0 + arr[n - 1];

And remember that n is 1, so:

5:    return 0 + arr[0];

Since arr[0] is equals to 2:

5:    return 0 + 2;

And then why sum([2, 3, 4], 1) returns 2.

I'm not sure if it's clearer now, but I hope so. :)

function operation() {
    console.log("testing");
}

function repeat(operation, num) {
    if (num === 0) return;
    operation();
    repeat(operation, num-1);
}

//repeat(operation, 10);
module.exports = repeat