您只需要表别名：

SELECT s.StoreID, s.StoreName, le.EmployeeName as LeadName,
       ae.EmployeeName as AssistantName
FROM s.Stores JOIN
     Employees le
     ON le.EmployeeID = s.LeadCashierID JOIN
     Employees ae
     ON ae.EmployeeID = s.AssistentCashierID;

提示：使用表别名的表缩写。它们使查询更容易理解。

可以使用表别名。

$sql = "SELECT StoreID, StoreName, leadCashier.EmployeeName as 
LeadCashier, 
assCashier.EmployeeName as AssistantCashier 
FROM Stores
JOIN Employees leadCashier ON leadCashier.EmployeeID = 
Stores.LeadCashierID 
JOIN Employees assCashier ON assCashier.EmployeeID = 
Stores.AssistantCashierID";
$result = $db->query($sql);

echo "<table>";
echo "<tr><th>Store ID</th><th>Store Name</th><th>Lead Cashier</th> 
<th>Assistant Cashier</th></tr>";

if ($result->num_rows > 0) {
// output data of each row
    while($row = $result->fetch_assoc()) {
    echo "<tr>";
    echo "<td>" . $row["StoreID"]. "</td>";
    echo "<td>" . $row["StoreName"]. "</td>";
    echo "<td>" . $row["LeadCashier"]. "</td>";
    echo "<td>" . $row["AssistantCashier"]. "</td>";
    echo "</tr>";
    }
} else {
    echo "0 results";
}
echo "</table>";