Correct way to save model loaded via the ViewModel links

What's the correct way to save a model that was loaded via the ViewModel links feature? Here's my ViewModel:

Code:

Ext.define('App.view.ClientFormViewModel', {
    extend: 'Ext.app.ViewModel',
    alias: 'viewmodel.clientform',
    links: {
        theClient: {
            type: 'App.model.ClientInfo',
            id: 1
        }
    }
});

This successfully loads the client with ID of 1 from the server, so after changing values, how do I get a reference to this model so that I can then save it? In my ViewController, I have this (so far, so that I can figure out the best way to go):

Code:

Ext.define('App.view.ClientFormViewController', {
    extend: 'Ext.app.ViewController',
    alias: 'controller.clientform',
    onSaveClick: function(button, e, eOpts) {
        console.log(this.getViewModel());
    }
});

I can see in FireBug that there are these possibilities:

The .data object which has a "theClient" property (i.e., this.getViewModel().data.theClient), but there is also a .linkData property which has "theClient" also.

The docs for .data say this:

This object holds the arbitrary data that populates the ViewModel and is then available for binding.

Which has me wondering if I should be referring to this.getViewModel().data.theClient object (the "arbitrary" is sewing seeds of doubt). Strangely enough, this.getViewModel().getLinks() is null.

Correct way to save model loaded via the ViewModel links