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 R1, and drops through resistor R3, if the reference direction for R3 is seen as positive at point A and negative at point B. The voltage rise through the power supply V1 can be described as a negative voltage drop, so V1 appears as a negative term.
I note that there's a discrepancy between the drawing and the equation you posted,
Bennington:
-V+Vr1+Vr3=0
Each resistor in the drawing has a polarity marker indicating the reference direction for the voltage across it. R3's polarity marker is at point B. Traveling aroung loop 1, the voltage across R3 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:
-V1 + VR2 - VR3 = 0
The sign of VR3 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 R3 * (I2 - I1). I1 appears as a negative term because the reference direction for I1 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