Thevenin Norton Concept
If you short those terminals, then the circuit will look like this:



R1 and R2 are now parallel. Hence current \$I_s\$ will divide between them. Therefore \$I_{sc} = I_s \frac{R_1}{R_1+R_2} = 5.6A\$. It is the current thru R2. 

The curren thru R1 will be \$I_s-I_{sc} = 8 - 5.6 = 2.4A\$

Mesh analysis:

Your KVL equation is wrong. Since current sources are there, you can use supermesh analysis by simply rearranging the circuit like this:



Supermesh equations:

$$14I_0 + 6I_1 = 0$$
$$I_1 - I_0 = 8 mA$$

On solving you get: \$I_0 = -2.4A\$ and \$I_1 = 5.6A\$

Note that \$I_1\$ is the same as \$I_{sc}\$ 

So we got the same results thru mesh analysis.