How do you find the current flowing through a short circuit?

That is easy. If there is a short around the resistor, then all of the current takes the path of least resistance, and bypasses the resistor completely. If there is no current flowing through the resistor, and Kirchhoff's Current Law (and the low of the Conversation of Energy) states than energy (in this case, the energy of the flow electric current) is never created nor destroyed, so all current flowing in must be equal to the current flowing out, then "Is" has to be equal to "It", unless I am missing something here. "It" has to go somewhere, and it isn't flowing through the resistor.

Without too much hand-waving, you can apply Kirchoff's laws here. By shorting out the resistor, you have essentially reduced the entire schematic to a single node. The sum of currents entering a node must be zero. Since you have 4A flowing into the node at the top, you must have 4A flowing out the bottom. Because the resistor has both terminals connected at the same node, they are at equal potentials and no current is flowing through it.

If you wanted to treat the circuit as two nodes separated by R and the "short," you would have to apply a practical resistance value to the short. You can certainly do this:

Give Rs a very small value, like 0.02 ohms. In practice it will be much less. Now consider that

Ir = 1/R (V1-V2)
Is = 1/Rs (V1-V2)
Ir+Is = 4A

50 (V1-V2) + 0.5 (V1-V2) = 4A

The current in the short is 100 times that in the resistor. If you are working in two significant figures, you have already reduced the current in R to zero. And the smaller you make Rs, the less current flows through the resistor.