Does a voltage source/diode in a circuit act as an open or a short circuit?
"According to the lecture notes, the diode acts as a battery after attaining Vf(forward bias voltage)." The model should have gone on to say if it does not get to forward bias it acts as an open circuit (reverse biased) so your 4.4V source is not connected. If the diode is reverse biased. 

You can determine if it is in reverse bias by looking at the current through the diode if current could flow  from cathode to anode it is reverse biased.

In your case this occurs if the voltage created by the R1,R2 divider is less than 5V in this case you have an open circuit instead of a diode. If the undefined voltage source is high enough to drive the diode into forward bias you have the case in your second diagram.