Kirhoff's voltage law is sometimes rendered as, "the sum of the voltage drops around a loop is zero." For a bit of charge starting at point B and going around loop 1, the voltage rises through the power supply, drops through resistor R
1, and drops through resistor R
3, if the reference direction for R
3 is seen as positive at point A and negative at point B. The voltage rise through the power supply V
1 can be described as a negative voltage drop, so V
1 appears as a negative term.
I note that there's a discrepancy between the drawing and the equation you posted,
-V+Vr1+Vr3=0
Each resistor in the drawing has a polarity marker indicating the reference direction for the voltage across it. R
3's polarity marker is at point B. Traveling aroung loop 1, the voltage across R
3 comes up as a voltage rise rather than a voltage drop, so I'd expect to see its appearance in the equation as negative, like this:
-V
1 + V
R2 - V
R3 = 0
The sign of V
R3 works itself out by how the current is represented: The voltage across the resistor is the resistance multiplied by the current entering the polarity mark. In this case, that's R
3 * (I
2 - I
1). I
1 appears as a negative term because the reference direction for I
1 shown in the drawing enters the other end of the resistor. That's the opposite of the signs of the two loop currents shown in your post:
Vr3=(I1-I2)R3
so, the math in your post is consistent. It's just not entirely consistent with the reference directions for voltage shown in the drawing.
Edit: clarity