The three diagrams in your first post are equivalent. The top resistor is connected across the top battery, and the bottom resistor is connected across the bottom battery.
Diagram B shows this the most clearly. The fact that the two circuits are connected together at one point doesn't affect either of them, as long as each one of them is self-contained, as you've shown them, and not connected to anything else.
Diagram C is interesting if you consider the current flowing through the section of wire in the middle, between the two resistors. This wire is part of both circuits, and will actually have the current from both circuits flowing in it.
I'll use conventional current in this explanation.
Current leaves the positive terminal of the top battery, flows through the top resistor, flows leftwards through the shared wire, and returns to the negative terminal of the top battery.
Current leaves the positive terminal of the bottom battery, flows leftwards through the shared wire, flows through the bottom resistor, and returns to the negative terminal of the bottom battery.
If you cut that wire in the centre, and put a current meter across the break, you would see the sum of the two circuit currents flowing in it.
Apart from that interesting fact, there is nothing special about any of those circuits. They can all be described as two batteries, each with a resistor load connected across it, with the two circuits joined at one point.
Your second set of diagrams is a different story. Although the two circuits look similar (the only difference is the polarity of the V2 battery), they are quite different. I will describe them separately.
The first circuit has two batteries with their negative sides connected together. Since they are the same voltage, their positive ends will also be at the same voltage; that is, +10V relative to the negative terminals.
For this diagram, I will refer all voltages to the negative terminals of the batteries. You have shown a ground connected to the positive terminal of V2; this causes confusion because this is not the natural "common" or "0V" or "reference" point of this circuit. This circuit could (and should) be redrawn in a way that makes it much clearer.
So the positive terminals of the two batteries are at exactly the same potential. You can connect them together, and nothing in the diagram will change. Effectively you have a single +10V supply feeding R1 from above, AND R2 from below. R1 and R2 are commoned and feed R3, which returns to the 0V point. (Remember I'm using the common negatives as the 0V reference point.)
R1 and R2 are effectively in parallel, and can be replaced by a single 500 ohm resistor supplied from a +10V source. This resistor and R3 are connected in series, across the 10V source. The total resistance is 1500 ohms, and from Ohm's Law, I=V/R where V=10 and R=1500 so I=0.006667 amps, or 6.667 mA, as shown by your simulator.
If you redraw that diagram with the common negative rail along the bottom and both batteries shown with their positive terminals at the top, you should be able to see what you have.
There are two separate 10V sources, which both have the same voltage because their negative sides are commoned. Each one of these sources feeds a1k resistor. This is equivalent to a single 10V source feeding two 1k resistors in parallel, i.e. a 500 ohm resistor.
This 500 ohm resistor feeds R3, which is returned to 0V. So the total resistance is 1500 ohms and the current is 6.6667 mA.
Half of this current comes from V1 and flows through R1, and half comes from V1 and flows through R2. Your simulation also shows this quite clearly.
The second diagram in that post is a different story. You have the batteries connected in series, so their voltages add together. You can see that the top rail voltage is 20V relative to the bottom rail.
(For this diagram I will refer all voltages to the ground rail that you have used.)
So you have a 0V rail (GND_0), a +10V point, and a +20V point. Across the 20V supply you have connected two equal resistors, R1 and R2. Because their values are equal, they will split the voltage equally; you will get 10V across R1 and 10V across R2.
This is all true regardless of the presence of R3.
You also have a +10V point coming from the centre tap of the batteries. This point is at the same voltage as the centre tap of the R1-R2 voltage divider. Since there is no potential difference across R3, no current will flow in it.
So to summarise the second diagram, you have a centre-tapped 20V DC supply feeding a voltage divider with two equal resistors, which provides a centre voltage as well. Connecting R3 has no effect on the circuit, since the voltage at both ends of R3 is the same and no current will flow through it.
