The .map() function is calling the Array() function with three arguments, the value of the array element which is 0, the index of that element, also 0, and a reference to the whole array.

So it's like doing this:

var a = [0];
var index = 0
Array(a[index], index, a);   // create array with three elements

The array returned by Array() then becomes the first element of the array that .map() creates, hence the extra level of nesting in your [[0, 0, [0]]] result.

EDIT regarding your edit: when you say [0].map(String); that results in String() being called with the same three arguments like String(a[index], index, a), but the String() function ignores all but the first argument, whereas Array() uses all supplied arguments.

Firstly, Array could be used as a function to create arrays:

var arr = Array(1, 2, "Hello");

console.log(arr); // [1, 2, "Hello"]

Secondly, map passes three parameters to its callback: the element, its index from the array and the array itself.

So, since your array contains one element, the line:

[0].map(Array);

is equivalent to:

[Array(0, 0, [0])];     // the element 0 in the original array will be mapped into Array(0, 0, [0])

After you updated question. Other answers provide you info about map

To answer why array and string differ look to constructors

String constructor accept 1 parameter String(thing) while array new Array(element0, element1[, ...[, elementN]])