使用PHP / AJAX将数据插入MySQL数据库，在插入后执行成功选项（回调）

I've been trying to make a simple site, and I can't quite wrap my head around some of the things said here, some of which are also unrelated to my situation.

The site has a form with 3 input boxes, a button, and a list. The info is submitted through a separate PHP file to a MySQL database, once the submit button is clicked. I'm supposed to make the list (it's inside a div) update once the info is successfully sent and updated in the database. So far I've made it work with async:false but I'm not supposed to, because of society.

Without this (bad) option, the list doesn't load after submitting the info, because (I assume) the method is executed past it, since it doesn't wait for it to finish.

What do I exactly have to do in "success:" to make it work? (Or, I've read something about .done() within the $.ajax clause, but I'm not sure how to make it work.)

What's the callback supposed to be like? I've never done it before and I can get really disoriented with the results here because each case is slightly different.

function save() {
    var name = document.getElementById('name');
    var email = document.getElementById('email');
    var telephone = document.getElementById('telephone');
    
    $.ajax({
        url: "save.php",
        method: "POST",
        data: { name: name.value, email: email.value, telephone: telephone.value },
        success: $("List").load(" List")
    });
}

Thank you in advanced and if I need include further info don't hesitate to ask.

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