Update a php variable with AJAX Javascript

You don't actually want to transfer the variable from BAConsultRecordsAJAX.php to BAConsult.php, but to BAConsult.js (I suppose that's the name of your JS file).

In reality, even that's what you wanted to do, it's impossible, because your main PHP is processed before the page even loads. What you can do, however, is overwrite the rendered value with a new one with the use of JavaScript.

To do that, send an AJAX request to BAConsultRecordsAJAX.php requesting the variable's value as shown below:

In JavaScript:

var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {

// Check if the AJAX request was successful
if (xhttp.readyState === 4 && xhttp.status === 200) {
    var td = document.getElementById("your table td value's id");
    var td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(""); // You don't need to send anything to the PHP file

In PHP:

<?php
     echo $skincareinuse;
?>

[EDIT]:

If you are using inline JavaScript use the following code:

<script type = "application/javascript">
    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function() {

    // Check if the AJAX request was successful
    if (xhttp.readyState === 4 && xhttp.status === 200) {
        var td = document.getElementById("your table td value's id");
        td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
    xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
    xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhttp.send(""); // You don't need to send anything to the PHP file
</script>

[EDIT 2]:

To pass a JSON object to PHP file via an AJAX request you have to make it a string. That requires the following code:

var JSONstring = JSON.stringify(yourJSONobject);

Then you can pass it in your PHP file with the AJAX request by putting it inside send() as shown:

xhttp.send("json_object=" + JSONstring);

Then in PHP, in order to use it, you have to decode it:

<?php
    $json_object = json_decode($_POST["json_object"]);
    // Now it's ready to use
?>

[About the code]:

Notes about the first file:

1. First of all, you have put your plain JavaScript code inside a PHP file and therefore it will not work. You have to wrap it in <script type = "application/javascript></script>

2. I don't have a clue what you are trying to do here:

Code:

var a = JSON.parse($(xmlhttp.responseText).filter('#arrayoutput').html());

• It seems like you are trying to filter out and parse the innerHTML of #arrayoutput.
• If the response is a JSON string, not HTML, so you absolutely can't do that. The logic of this above line is flawed.

How I would write your code:

function showconsultationdata(str) {
    var xmlhttp;
    if (!str) {
        $("#txtHint2").empty();
    } else {
        xmlhttp = new XMLHttpRequest();
        // Providing support for a 15-year-old browser in 2016 is unnecessary

        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                $("#txtHint2").html(xmlhttp.responseText);                       
                var a = JSON.parse(xmlhttp.responseText);
                $("textarea#skinconditionremarks").val(a.skinconditionremarks);
                $("textarea#skincareremarks").val(a.skincareremarks);

                var test = a.skincareinuse; // The variable you want
                // Its saved in "text" and can use something like:
                // $("#element").html(test); to have it replace your current text
            }
        };
        xmlhttp.open("GET","BAConsultRecordsAJAX.php?q="+str,true);
        xmlhttp.send();        
    }
}

Notes about the second file:

1. I don't know if the query works for you, but it probably shouldn't, because:
   • $_SESSION is an associative array and you pass nric, whereas it should be "nric".
   • As $_SESSION is an array, the proper method to insert its value to your query is: {$_SESSION["nric"]}.
2. To ensure that you don't become the victim of an SQL Injection, use parameterised queries or at least sanitise somehow the received data, because GET is relatively easy to hack. Check the improved code later on how to do it.

How I would write your code:

$q = $_GET['q']; //get dateconsulted value

$dbconn = mysqli_connect("localhost", "root", "") or die("Error!");
mysqli_select_db($dbconn, "database") or die("Error!");

$consult = "SELECT * FROM Counsel where nric='{$_SESSION["nric"]}' and dateconsulted = ?";

if ($stmt = mysqli_prepare($dbconn, $consult)) { // By using a prepared statement
    mysqli_stmt_bind_param($stmt, "s", $q);      // you eliminate the chances to have
    if (mysqli_stmt_execute($stmt)) {            // an SQL Injection break your database
        $consultresult = mysqli_stmt_get_result($stmt);            
        if (mysqli_num_rows($consultresult) > 0) {
            while ($row = mysqli_fetch_array($consultresult)) {
                $skincareinuse = explode(",",$row['skincarecurrentlyinuse']);
                $skincondition = explode(",",$row['skincondition']);
                $queryResult[] = $row['skincareremarks'];
                $queryResult[] = $row['skinconditionremarks'];
            }
            $skincareremarks = $queryResult[0];
            $skinconditionremarks = $queryResult[1];
            $array = array('skincareremarks'=>$skincareremarks
                           'skinconditionremarks'=>$skinconditionremarks
                           'skincareinuse'=>$skincareinuse,
                           'skincondition'=>$skincondition);
            echo "<div id='arrayoutput'>";
            echo json_encode($array);
            echo "</div>";
            mysqli_stmt_close($stmt);
            mysqli_close($dbconn);
            exit;
        }
    }
}

Obviously you may change the values to suit your requirements, you must have jquery installed, the time value could be anything you want

$.ajaxSetup({ cache: false }); 
setInterval(function() { 

var url = "pagethatrequiresrefreshing.php";
$('#divtorefresh').load(url);
 }, 4000); 

Using include will most likely work. From the docs:

When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward.

BAConsultRecordsAJAX.php

while($row = mysqli_fetch_array($consultresult)) {
    $skincareinuse[] = explode(",",$row['skincarecurrentlyinuse']);
}

include "BAConsult.php";