Does this mean the voltage going over R3 is supplied solely by B2?
We know the Thevenin voltage is 11.2V and we know that the internal resistance is 0.8 Ohms. So, we add in the 2 Ohm load resistor and we get the current through the center resistor
11.2V / (2.8Ohms) = 4 Amps. So, 4 amps flows down the center resistor, out of the node. Working back up through the 2 Ohm load resistor, we find that the center point is 8V (4A * 2 Ohms).
If the center point is 8V and B2 is 7V then there is 1V dropped across R3 (we have to get from 8 down to 7) so the current flow is INTO the battery - it is getting charged! The current flow is out of the node.
Well, we know that there is 1A flowing to the right out of the center node toward the battery and we know there is 4A flowing out of the center node down through R2 so there must be 5A coming into the center node through R1. That means the drop across R1 is 20V (5A * 4Ohms) which means the center point should be 28V - 20V (=8V) and that's good because we already established that point.
The circuit balances!
The current through R1 is 5A going into the center node.
The current through R3 is -1A going into the center node (which means it is actually flowing out of the node)
The current through R2 is -4A going into the center node (again, the current is flowing out)
The sum of the currents going into and leaving the center node is zero - good thing!
This nodal analysis is the easy way to figure out how the circuit works without resorting to Thevenin.
The current going into the node from the left is (28-V1)/4
The current going into the node from the bottom is V1 / 2
The current going into the node from the right is (7-V1)/1
Realizing that the sum of the currents going into the node must equal zero, we just add them up. What we wind up with is the node voltage and from there it is easy sailing. Note that the current from the right will be negative as will the current coming up from the bottom. This is a sign convention issue. We assumed the current was flowing into the node, the minus sign just says we were wrong and the current is actually flowing out of the node.
The equation for the nodal analysis is
(28-V1)/4 + (7-V1)/1 + (0-V1)/2=0 Note how the 3rd term is convoluted to force current INTO the node.
28-V1+28-4*V1-2*V1=0 -- multiply through by 4
56-7*V1=0
V1=8V
It seems silly to assume the load current flows into the node and you will get the same answer if you assume it flows out because that current will be subtracting (V1/2) which is just what happened above!
Just be consistent in how you draw the arrows and write the matching equations. Everything will work out!