I need to find the voltage drop between R1 (4 Ohm) and R2 (2 Ohm). I tried applying Kirchhoff's law in this way: \$-10+4I-8-2I=0\$ and then solving by \$I\$ I have \$I=9 A\$ but that's not correct. Can someone help me?
Why did you change the sign for the voltage sources and for the resistor drops? Imagine first there was only one source and one resistor (R10). You would have +10 - 4I = 0.
Now add the second resistor, you get +10 - 4I - 2I = 0. If you aren't comfortable with that, imagine first combining both resistors into one, it would be 6 Ohms, right? So you would have +10 - 6I = 0, which is the same as above.
Per KVL, the order of the the components doesn't matter, so you can apply the same logic to the sources. A 10V and 8V source in series and in the same polarity produce a 18V source. Don't be fooled by the fact the 8V source is on the right side of the circuit, upside-down; it's still adds to the total voltage source. So you would have +10 + 8 -4I - 2I = 18 - 6I = 0
You might find it easier to rearrange the drawing like this:
Remember though this doesn't hold once you add any parallel item or branched circuit.