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
To determine whether friction is static or kinetic, examine whether the two surfaces are moving relative to each other -- if so, the friction is kinetic, and if not, it's static. It doesn't matter which inertial reference frame is chosen. Even though the ground is an inertial frame, the relative motion between the box and ground means that friction is kinetic here.
This makes sense in the context of the physical difference between kinetic and static friction. Static friction usually results from two surfaces "interlocking" together into a configuration that requires additional force to change. On the other hand, kinetic friction usually results from chemical interactions between the two surfaces. The interlocking can only occur between surfaces that aren't in relative motion.