Static friction :
Static friction opposes relative motion between two surfaces. The car is moving at some tangential velocity relative to the road. Thus, static friction should act in the opposite direction. Yet it acts down the incline rather than parallel to the motion of the car. If it acts down the incline, it's essentially helping the car maintain the same relative velocity w.r.t the incline rather than opposing this relative motion.
Kinetic friction :
Consider a block placed on a long, rough plank which is moving to the right at some constant speed, and the block is moving to the right, but slower compared to the plank, so relative to the plank, the block is sliding towards the left. Will kinetic friction act to the left to reduce the kinetic energy of the block, or will it act towards the right to oppose the relative motion of the block and plank? If it acts to the right, doesn't that increase the kinetic energy of the block, which is something kinetic friction should not do?
Best Answer
Just addressing the question in this comment about banked curves. Static friction is always going to oppose the motion that would happen if there were no friction. I will use the free-body diagram here as a reference for the case of no friction. The only two forces on the car are the normal force (N) and gravity (mg). The sum of these two forces is in the horizontal direction toward the center of the circle that the car is traveling around. This net force is what keeps the car traveling in a circle, and is equal to a component of the normal force. Now, if we consider the fourth equation on that page, which comes from considering $F_{net}=F_{centripetal}$:
$$mg\tan{\theta} = \frac{mv^2}{r}$$ And divide by $m$: $$g\tan{\theta} = \frac{v^2}{r}$$
This equation says for the car to stay in uniform circular motion (speed $v$ and radius $r$ don't change), there must be a balance between the four parameters in this equation. If, for example, speed $v$ is increased, radius $r$ must also increase given that $g$ and $\theta$ are constant. In the case that the car starts increasing its speed, it will start to slide up the incline. In this case, it will do so, and stop sliding sideways once the equation above is satisfied.
However, if we consider the case of an incline with friction, the situation changes. First, if the equation above is satisfied, then no friction will act sideways on the car tires (it isn't necessary, the car isn't trying to move sideways). However, there will be some friction on the tires in the forward direction, as in any case of "rolling without slipping." That phrase means the point of the tire that is in contact with the road at any instant in time is not moving w.r.t the road. It is "trying" to move backwards (think about a car at rest on ice. If you try to accelerate too quickly, the tires spin, and the point on the ice moves backwards. It is the same with a car in motion.), so the force from static friction must push forwards on the tire. This is what allows the car to accelerate forwards.
In the same scenario (rolling without slipping, $g\tan{\theta} = \frac{v^2}{r}$ initially satisfied) if the car's speed increases, then the equation will no longer be satisfied. But there is friction now, and as we found in the frictionless case, the car will "try" to move up the incline, and thus static friction will point down to oppose this motion. Vice versa, if the car decreases its speed, static friction will point up to oppose the car "trying" to slide down. (In this way, you can drive at a range of speeds for a given $g$, $\theta$, and $r$ if static friction is present.)
So, static friction always opposes attempted motion between two surfaces in contact.