It seems like the issue lies with the 'options' problem here:
var dropdown = $('#cars');
dropdown.options = function(data, selectedKey) ...
This code works in JavaScript, but when using TypeScript, you might encounter an error due to static type-checking. Since dropdown is a jQuery object and jQuery type definitions do not declare an options property, assigning to it will result in an error. To work around this, you can typecast dropdown to any so that it behaves like a regular JavaScript object:
var dropdown: any = $('#cars');
dropdown.options = (...)
The same error occurs with .Default where you are initializing storedPreferences as:
var storedPreferences = localStorage.getItem(...);
Since TypeScript infers the return type of localStorage.getItem as string, it treats storedPreferences as a string. In TypeScript, you cannot change the type of a variable once declared, hence the error when trying to access a non-existent property like Default.
To solve this, you can explicitly declare storedPreferences as type any:
let storedPreferences: any = localStorage.getItem(...);
The main issue here is that you're coding in TypeScript as if it were JavaScript, leading to compiler errors. Remember that in TypeScript, you cannot add new properties to objects of known types without declaring them as any first. This way, TypeScript will behave more like JavaScript, but you lose some advantages of TypeScript.
Additionally, it's recommended to use let or const instead of var.
The code on jsfiddle functions correctly because it is written in JavaScript, not TypeScript.