function errMsg(code, msg) {
    const eCode = '<b>E-NS: ' + code + ' </b> </br>' + msg;
    const eMsg = '<span class="err" style="color: red;">' + eCode + '</span>';
    
    $('.notify').html(eMsg);
}

$(document).ready(function() {
    $("#form-2").hide();
    $('#next-1').click(function(e) {
        e.preventDefault();
        // console.log("this will be sent to the server:",$('#form-1').serializeArray());
            $.ajax({
                url: 'https://jsonplaceholder.typicode.com/posts', // '../data.php',
                method: 'POST',
                data:  $('#form-1').serializeArray(),
                dataType: 'JSON',
                success:function(response){
                    // console.log("this came back from the server:",response);
                    if(response){
                        $('#form-1').hide();
                        $('#form-2').show();
                        errMsg(response.id, JSON.stringify(response))
                    } else {
                        console.log(response.status);
                        console.log('Response = ' + response.code + response.error);
                        errMsg(response.code, response.error);
                    }
                }
            })
    })
})

.notify {border: 1px solid grey}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<form id="form-1">
First form ... <br>
<input type="text" name="title" value="my very own title"><br>
<input type="text" name="size" value="XL"><br>
<input type="number" name="quantity" value="3"><br>
<button id="next-1" type="button">OK</button>
</form>

<form id="form-2">
Second form ... <br>
<input type="text" name="title" value="some other product title"><br>
<input type="text" name="size" value="M"><br>
<input type="number" name="quantity" value="5"><br>
<button id="next-2" type="button">OK</button>
</form>

<div class="notify">This is the place for messages ...</div>

您还应该检查服务器将发回的内容。对于我的虚拟 JSON API，没有发回 .status 属性。所以对其进行测试没有意义。