<p>Hey Cheese,</p>
<p>all three images you've attached show potentiometer wired up as an attenuator. This means that only two lugs are used and that you are varying the resistance between the center lug and one of the others. The confusing part is that he has depicted this two different ways in the same schematic. This can be represented as<br />
A). a resistor with an arrow through the middle (not shown in this schematic but common) or<br />
B). as a resistor with an arrow pointing to the middle. There are two ways this can be drawn which are both shown in your images above. one is a resistor only connected at one end. the other is shown with the center lug connected directly to one of the other lugs. They are completely the same. as long as you wire the center and outer lug as indicated you're good. </p>
<p>"In this next pic, I understand that the negative lead of the cap goes to ground and the positive connects to position 1 of the 40106, but I was wondering if there is a difference between connecting both the positive terminal of the cap and the wire going to the pot (VR1) directly to the IC or if they should be separated like the drawing suggests."</p>
<p>don't take the schematics too literally... they simply indicated what needs to be connected to what. The fact that the cap and pot connect to one point then continue as one line to pin 1 doesn't mean you have to solder them together then put a jumper to pin 1 or anything like that. simply connect the cap and pot to pin 1 in whatever way is most effective. There's a lot of flexibility with this kind of circuit.
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