I answered to the other ticket opened on the same subject. I would like to ask you continue our discussion there to avoid any duplication. Thank you for the understanding.
Here is a quote of the answer:
In general, the DropDownList is design to post its selected value defined by dataValueField option. It behaves just like every HTML input element. If you would like to send more information to the server, then you will need to get the selected data item. Use the dataItem method for this task:
From here, you can send the data item using Ajax or posting with regular Html form element is required, then you will need to generate the required HTML input elements in order to post the additional data.
widget = $(
dataItem = widget.dataItem();
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