Yes, indeed while "potentiometer" nowadays means "variable resistor" - a fresh glance at the word itself will show that it really means "potential measuring device".
And that was its original purpose.
I have seen and used a real Potentiometer exactly once, at a six hundred year old university, as an exercise. (The lab equipment was much newer, and quite splendid!)
EDIT : the form of potentiometer described here is also known as a "metre bridge".
It is a length of resistance wire, stretched out along a board (mahogany, of course) over a metre ruler, with a very solid brass (what else!) bar alongside it, and a sliding right-angle knife-edge contact between the bar and the resistance wire. It is connected as your illustration shows, to the top cell, an unknown but steady voltage source (usually a 2V lead acid cell) with an unknown series resistance. The precise values of V,R don't matter.
The metre rule can be engraved on the brass bar, and with a Vernier scale on the sliding contact, the instrument can give four digit accuracy, better than many a digital multimeter today.
I believe K1 and K2 are brass tapered pegs which fit into tapered gaps in the brass bar, i.e. switches.
Now carefully connect up your unknown cell via G, which is preferably one of Lord Kelvin's fine mirror galvanometers, moving a focussed spot of light to amplify the smallest motion, and adjust the knife edge contact for zero current. At this point you have a length (measured from the ruler) representing the potential of the unknown cell.
Because you don't know the driving voltage, you also need to make the same measurement with a "standard cell" usually a Weston cell to get the length representing 1.01864 volts; the ratio between them gives you your actual cell voltage.
CAUTION : if you are careless in this step and draw significant current, you will wreck your valuable standard cell.
Now you can return to the unknown cell. Re-establish its open-circuit voltage (length) - any change shows that the driving cell needs to be recharged.
Then insert peg K2, connecting known resistance R2, and establish the new voltage (length) when driving current through R2.
The rest is mere arithmetic...
Given that you've said R7 is set to 0 ohms, the two fixed terminals of R6 are shorted together. When the wiper is at either extreme, its resistance to either fixed terminal will also be 0 ohms. When the resistance is 50%, both halves will be in parallel. Given that this appears to be a 50 k pot, we'd expect
$$R_{eq}=\frac{R_{max}}{2}||\frac{R_{max}}{2}=\frac{R_{max}}{4}=\frac{50000}{4}=12500$$
just as you measured.
Best Answer
Using the third leg turns the rheostat into a potentiometer. The whole thing acts as a voltage divider, with the moving wiper varying the ratio between the two resistances connected to the differing voltages.