the 2 upper schematics in your diagram, that these are the voltage shifter and would fit in with the voltage across the 10K resistor to produce the required 2 to 12V, please correct me if I am wrong.
Not sure what do you mean by the diodes being 2V ref voltage.
The two upper schematics are voltage sources that can sink current for the protection diodes so that the 2-12 V doesn't ever go outside that range.
The upper 10 K output pin can't go above 13 + 1 = 14 V and the lower one can't go below 1 - 1 = 0 V.
Some voltage sources can only source current; driving current into them makes their output voltage higher. These two sources can sink and source current.
A forward biased conventional diode (like a 1N4148 or 1N4001) can act like a ~1 V Zener diode
when you push rated current through them; that is, the V-I curve is almost horizontal [with V on the Y-axis and I on the X-axis] at around 1 V, but diodes used like this are not precise and are temperature sensitive.
So, two diodes in series
at rated current approximates a 2 V Zener. This is what shifts the 0-10 V output into 2-12 V out.
You can see by your meter that silicon diode D7 is barely turned on, because ~14 V/15 kΩ = ~1 mA, and that's not enough current to turn it on fully. You can reduce the 15 KΩ resistor to get more current and a higher Vdrop across that diode.
D2 and D3 are also not fully turned on and that's why you got only 11.3 V instead of 12 V. You could increase their current to 10 mA by running a 1300 Ω resistor from the anode of D2 up to +15 V. I considered putting in this "keep alive" resistor but I left it out for the sake of simplicity.
Note that the Vdrop across a diode is due to the Vdrop across the PN junction plus a small drop across the diode "bulk resistance", this resistance resulting from the material they used to make the diode.
I should have labelled all these diodes D1, D2, etc.
The one on top of the current source (your computer calls this D1) prevents the 2 V diode drop from pushing current back into the source through the 10 KΩ resistor. 2 V/10 kΩ is 0.2 mA so this would introduce a (100%)*(0.2/1) = 20% error in the current source output.
This should explain the purpose of all the components.
Revision A:
Change the 13 V Zener to 12 V; I just noticed this minute that when the output is at 12 V the protection diode D4 is turned on. Since it's not connected in your circuit, you didn't notice this effect.
I must say this thread has been fun, and I didn't expect SW would be breadboarding this back-of-an-envelope circuit.
How do you do the attached thumbnail that can be expanded? I hate posting mostly white paper.