A question related to threshold voltage in relation with the body effect in MOS
Here's how I would analyze the threshold voltage.  This is the short answer.  I can elaborate, so feel free to put suggestions in the comments.  I want to make a few notes first.
notes:
There are a few ways to define the threshold voltage that I have seen.  When you have access to the devices, you do not know the doping so you cannot know where the exact Fermi levels are.  When you are just pushing through the physics you make assumptions about what the dopings are, which differs in practice.  Now back to the "text book" approach:

I present to you an nFET that looks as follows:


In the image, \$\Phi_s\$ is the surface potential and \$\kappa\$ is the channel divider and this term is what relates to your graph, and I define this as
$$\kappa=\frac{C_{ox}}{C_{ox}+C_{dep}}$$
Deriving the threshold for the simplest condition from first first principles (no charge sharing on the sides), you should get the following (corrections welcome):

$$V_{T0}= V_{FB} + \frac{V_{BG}}{\kappa} +\frac{2 U_T}{\kappa} \ln\left[\frac{N_D}{\sqrt{N_C N_V}}\right],$$
where \$U_T\$ is the thermal voltage, \$V_{BG}\$ is the band-gap, and \$N_C\$ and \$N_V\$ are the density of states for the conduction and valence bands respectively, for temperature these are fixed.  \$V_{FB}\$ is the flat band condition and this will change depending on the "junk" in your oxide, and this is your lumped offset, which is the extra voltage you are eluding to.

The change in the divider,\$\kappa\$, to the surface is what is causing your threshold shift in the graph.  (I can also tell you that device was very small, probably a DIBL device to get behavior like you see in the graph. )