The Norton current is calculated by replacing the load with a short circuit and then figuring out how much current flows through the short circuit. There are two loops in the figure shown with the short. The total voltage drop around the loop must be equal to zero. Each loop only has two components, a source and a resistor. So all the voltage provided by the source only has one component to drop all its voltage across. So each resistor sees the full sources voltage across it. So VR1 = 28v and VR3 = 7v. Then using I=V/R, you can calculate the current through each resistor. IR1 = 28v / 4 = 7A, and IR3 = 7v / 1 = 7A.
Then, using KCL, the sum of currents entering a node must equal the sum of the currents leaving the node. There are 14A leaving the node, so there also must be 14A entering the node.
The 0.8 ohms comes from calculating the circuit's resistance as seen by the load. Use Superposition and remove the sources, voltage sources become shorts, current sources become open circuits. The resulting circuit is only 2 resistors in parallel. (1*4)/(1+4) = 0.8 ohms.
Another good way to find the Norton Resistance is to remove the load. Calculate the voltage open circuit...Voc. Then calculate current shorted across the load connections...Isc. Rt = Rn = Voc/Isc. Voltage open circuit can be found by using voltage division. The voltage across R1 and R3 is 28-7=21v. The voltage across R1 can be found by 21v * (1/(1+4)) = 4.2v. So the total voltage across the load connections equals 4.2v + 7v = 11.2v.
Rt = Rn = (11.2v / 14A) = 0.8 ohms
-Jacob