If we want to go to store A and store B, the idea is to park the car somewhere in between so we walk the least possible. Intuitively, parking right in the middle of the two stores is the best option, but a simple remark from Euclidean geometry tells us a different story.

Let us call the stores $A$ and $B$, and the car $C$ for simplicity. Hence what we want to minimize is $CA+AB+BC$, which is just $2AB$ if $C$ is in between $A$ and $B$, regardless of specific position of $C$. Therefore, it really doesn't matter where we park as long as it is in between $A$ and $B$.

If we consider the distance from the actual parking spot to the stores, we turn this into a planar problem

and again, our purpose is to minimize the distance $CA+AB+BC$, where now is the perimeter of a triangle with a fixed side

For this, a better approach is to consider the locus where the car gives a fixed total distance $CA+AB+BC$. This is a well known problem whose answer is nothing else but an ellipse whose major axis is $AB$.

Thus, In order to minimize the total distance that one would have to work, we have to look at all the parking spots inside the elliptical regions and find the one with smallest distance to the major axis, i.e., the closest parking spot that is in the middle of $A$ and $B$. Hence, as our intuition would tell, parking in the middle is the best strategy if we allow jaywalking in the parking lot.

If we want to be a little bit more 'rule follower' (pun intended) and we walk only in rectangular coordinates, the problem reduces to the one dimensional version of it and where we park actually wouldn't matter as long as it is between $A$ and $B$.