There are rules, such as the passive sign convention, and there are other rules (which shall remain unnamed as I don't like them.)
So long as you use logically consistent rules and apply them appropriately to a valid circuit graph with one and only one solution then the results are always the same, regardless of the convention.
Your circuit shows the application of the passive sign convention, where the resistor's (+) side occurs where the current arrow shows the current entering the device and where the resistor's (-) side occurs where the current is shown exiting the device.
To apply KVL for a series loop like the one shown in your schematic, you start at any node in the loop and call that place \$0\:\text{V}\$. (Yes, that's arbitrary. But you are always allowed to call one and only one node as the reference point, or \$0\:\text{V}\$.) Then you can go around the loop in either direction -- following the current arrow or going against it, doesn't matter which -- and total up the results by always looking at the sign on the side of the 2-terminal device opposite to the node you are currently at.
I've taken the liberty of labeling all three nodes in your circuit so that we can communicate better:
There are three possible starting points and two possible directions around the loop. That means there are six ways of writing down the exact same thing:
\$\begin{align*}0\:\text{V}+V_{\small{S}}-I\cdot R_1-I\cdot R_2&=0\:\text{V}\tag{CW from N1}\\\\0\:\text{V}-I\cdot R_1-I\cdot R_2+V_{\small{S}}&=0\:\text{V}\tag{CW from N2}\\\\0\:\text{V}-I\cdot R_2+V_{\small{S}}-I\cdot R_1&=0\:\text{V}\tag{CW from N3}\\\\0\:\text{V}+I\cdot R_2+I\cdot R_1-V_{\small{S}}&=0\:\text{V}\tag{CCW from N1}\\\\0\:\text{V}-V_{\small{S}}+I\cdot R_2+I\cdot R_1&=0\:\text{V}\tag{CCW from N2}\\\\0\:\text{V}+I\cdot R_1-V_{\small{S}}+I\cdot R_2&=0\:\text{V}\tag{CCW from N3}\end{align*}\$
All of the above result in the exact same equation: \$V_{\small{S}}=I\cdot\left(R_1+R_2\right)\$.
In the above \$I\$ will be positive since the assumed current direction is the same as the actual current direction in that circuit.
Now you can reverse the assumed loop current direction so that goes CCW instead of CW. Per this:
This will set up six more possible ways of writing out the same thing:
\$\begin{align*}0\:\text{V}+V_{\small{S}}+I\cdot R_1+I\cdot R_2&=0\:\text{V}\tag{CW from N1}\\\\0\:\text{V}+I\cdot R_1+I\cdot R_2+V_{\small{S}}&=0\:\text{V}\tag{CW from N2}\\\\0\:\text{V}+I\cdot R_2+V_{\small{S}}+I\cdot R_1&=0\:\text{V}\tag{CW from N3}\\\\0\:\text{V}-I\cdot R_2-I\cdot R_1-V_{\small{S}}&=0\:\text{V}\tag{CCW from N1}\\\\0\:\text{V}-V_{\small{S}}-I\cdot R_2-I\cdot R_1&=0\:\text{V}\tag{CCW from N2}\\\\0\:\text{V}-I\cdot R_1-V_{\small{S}}-I\cdot R_2&=0\:\text{V}\tag{CCW from N3}\end{align*}\$
These six result in the exact same equation: \$-V_{\small{S}}=I\cdot\left(R_1+R_2\right)\$. However, this equation is different from the earlier one, with \$I\$ being negative this time, since the assumed current direction is opposite to the actual current direction in that circuit. (The fact that it is negative is what tells you that you got the assumed direction wrong.)
Now, there are other conventions which will just mean that there are still more ways to diagram the circuit and come up with some result. But no matter what you do or how you do it, so long as the method is logically consistent and applied correctly, the final results may be interpreted identically.
Finally, if you want to learn to apply a specific convention well, you will need a much more complex circuit than this one to make sure you have mastered it.