Here is an example how to use form onsubmit method

Your HTML part

<form id="formTest" method="post" enctype="multipart/form-data">

        <label for="file">Filename:</label>
        <input type="file" name="files" id="files" />

        <input type="submit" name="submit" value="Upload" />
    </form>

script

<script type="text/javascript">
var form = document.getElementById('formTest').onsubmit = function (e) {
        e.preventDefault();
    var formdata = new FormData(); //FormData object
    var fileInput = document.getElementById('files');
    if (fileInput != "" && fileInput.files.length > 0) {
        //Iterating through each files selected in fileInput
        for (i = 0; i < fileInput.files.length; i++) {
            //Appending each file to FormData object
            formdata.append(fileInput.files[i].name, fileInput.files[i]);
        }
        //Creating an XMLHttpRequest and sending
        var xhr = new XMLHttpRequest();

           var url = '@Url.Action("Index","Home")';
        xhr.open('POST', url);
        xhr.send(formdata);
        xhr.onreadystatechange = function () {
            if (xhr.readyState == 4 && xhr.status == 200) {
                var result = xhr.responseText;

            }
        }
        return false;
    }
}
</script>

C#

public ActionResult Index()
{ 
   if (Request.Files.Count > 0)
   {
      var file = Request.Files[0];    
      if (file != null && file.ContentLength > 0)
      {
         var fileName = Path.GetFileName(file.FileName);
         var path = Path.Combine(Server.MapPath("~/Images/"), fileName);
         file.SaveAs(path);
      }
    return View();

   }
}