I understand now, that the static friction opposes the tendency of a surface to slip over another.
Exactly! Let's remember this point in the next sentences.
When a wheel rotates, static friction pushes the wheel forward to keep the contact surface stationary with respect to the ground.
Not necessarily forward. It can also be backwards. It pulls in whatever direction necessary to keep the contact point stationary. And only if necessary (e.g. there is no static friction when a wheel rolls at constant speed over a horizontal surface. There are no forces at all present, so nothing for the static friction to oppose. The rolling motion just continues effortless until stopped.)
When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too.
No. It pulls uphill.
Think for a moment of a star rolling down. It's legs touches the ground one at a time. While a leg touches the ground, it must not slide. We can think of it as a stationary object in that moment while it is touching. Downwards forces (gravity) must be opposed by static friction upwards, just like for a stationary object resting on the hill.
Then the next leg takes over, and the same thing is the case. Static friction must hold back upwards to avoid sliding of the leg.
Now add more legs to the star. Many, many more. Soon you almost have a round circle, where the legs are the "points" of the circle that touch the ground for only a split-second. Nevertheless, the same is the case; while touching, static friction holds on to the touching point upwards to avoid it from sliding because of gravity.
[...] for the car to go up the hill, where the static friction is again pushing the car forward, countering the slipping of the tires that would result due to the rotation of the tires.
Yes. Again static friction pulls uphill to prevent sliding because gravity pulls down. The direction of static friction does not depend on the rolling direction; it doesn't care if you roll up or down the hill.
If the wheel accelerates at the same time, the static friction might point differently. This is again regardless of the rolling direction but only the acceleration direction comes into play.
How can you get a constant velocity when going down hill, when the static friction and gravity are both creating a net force down the hill (if you ignore rolling friction)?
You are exactly on point here. It can't have constant speed, if all forces pull the same way. Such thinking will lead to the understanding that static friction in fact must point the other way.
A question asks me to find the steepest hill a car can descend at constant speed given the static friction coefficient, but I think what is needed is the rolling friction coefficient. I don't think its possible to answer the question without rolling friction.
As mentioned in a comment, most questions would assume ideal world-models. No deformation of surfaces e.g. So rolling friction will be assumed 0 most often, unless you drive on a clearly non-ideal surface, like a sandy beach or a flexible trampoline.
The idea to solve this question is to remember the formula for maximum static friction:
$$f_s\leq \mu_s n$$
If you have a certain friction coefficient $\mu_s$, you can do your Newtons 2nd law calculations on the car and put this formula in with an $=$ instead of $\leq$, because you are looking for the maximum limit. Then solve it for the slope angle (the angle will be a part of the force components).
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.
Best Answer
The problem with your argument lies with this:
That isn't quite correct. When the wheel is spinning up, then yes, static friction is what accelerates it, but once it is rotating, no torque is required to keep the rotation going. So once the wheel is moving at some constant velocity, there is no static friction, and in fact no external force at all, thus there are no changes in energy.
In reality, every wheel has some amount of dissipative force acting on it, usually due to deformation. The collective effect of the dissipative forces acting on a rolling wheel is sometimes called "rolling friction," and it acts on the wheel in much the same way as normal friction: it removes energy. Given the constraint that $v = \omega r$, the energy of a wheel is equal to
$$E = \frac{1}{2}(I + mr^2)\omega^2 = \frac{1}{2}\biggl(\frac{I}{r^2} + m\biggr)v^2$$
so when energy is removed, both the angular and linear speeds will decrease. This is what causes the wheel to slow down and eventually stop. Note that in this case, there will be some nonzero amount of static friction as required to prevent slipping.