Here's how it looks with a lot of the shorthand expanded:

const reducer = (accumulator, currentValue, index) => {
  if (currentValue > accumulator[1]) {
    return([index, currentValue])
  } else {
    return(accumulator)
  }
}

let largest = accumulator.reduce(reducer, [-1, 0])

What's happening:

Array.prototype.reduce() has two arguments, the callback function and the initial value. That's where the [-1, 0] is coming from, that's the "initial value" that will be passed as the "accumulator" to the callback function on the first iteration.

The shortest shorthand for a function in javascript is () => value where "value" represents the return value of the function—no parenthesis are required. In this example that return value is a ternary statement: condition ? return if true : return if false. Essentially, because they're using a single line to produce a single value, there's no parenthesis or braces required around the function (which ends at the a).

The callback function you pass to Array.reduce() can take three arguments, (accumulator, currentValue, index).

The key here is in the initial value, which is [index, value], which keeps track of both until iteration is finished, in which case just the final value is returned.

The array that comes after the comma is a second argument to .reduce(). The code there is equivalent to

let largest = arr.reduce(
  (a,v,i) => {
    return v > a[1] ? [i,v] : a;
  },
  [-1,0] // second argument
);

On very iteration, the accumulator is an array composed of 2 values. The value returned from the callback becomes the new accumulator in the next iteration. The value returned on the last iteration becomes the result of the whole .reduce() expression - the largest.

So what return v > a[1] ? [i,v] : a; does inside a .reduce() callback is it'll go through the array and find the item with the highest v - once the [i, v] of that becomes the accumulator, the accumulator will stay as that array until iteration finishes.

This is more easily understood if formalized and expanded:

let reducer = (a,v,i) => {
  if (v > a[1])
    return [i,v];

  return a
};

let largest = arr.reduce(reducer, [-1,0]);

Where now the reducer function argument is pretty clear.