About The operator of javascript (|| ,&&)

I am reading the source code of underscore.js, then something confused me;

  // Its code, check the passed-in parameter obj 
  _.isObject = function(obj) {
    var type = typeof obj;
    return type === 'function' || type === 'object' && !!obj;
  };

I am confused about the operator order of expression;

I think the operator precedence in

return type === 'function' || type === 'object' && !!obj;

will be from left to right; I mean equal to :

return (type === 'function' ) || ( type === 'object' && !!obj);

if type equal function return true; else oprate type === 'object' && !!obj ;if type equal object return !!obj ,same as Boolean(obj); else return false;

I made some examples:

var a = alert(1) || alert(2) && alert(3);
alert(a); //result : 1, 2 undefined;


var a = alert(1) || alert(2) && 0;
alert(a); //result : 1, 2 undefined;

what confused me:

1. why !!obj should exsit ?if we delete !!obj ,code run as well;

2. the operator order of this code? I know && oprator are higher than ||, so I guess !!obj effect when obj is null; but when I pratice that is no what I want;