The article doesn't go into detail on curvature. A fast-moving asteroid would curve slightly towards the Earth due to gravity, but the bouncing off the atmosphere would be in the opposite direction, tending to cancel out the effect of gravity (at the expense of some kinetic energy). We could thus picture the object's path as being linear, especially at higher rates of speed. If we graphed the object's height above the Earth's surface vs. time, initially there would be a steep decline as the object approaches, a flattening as the object passes parallel and closest to the surface, and then a steepening incline as it moves away again.
If we consider that the atmospheric pressure decreases exponentially with height, we can see that the vast majority of the interaction with the atmosphere is going to occur in the relatively small portion of the graph where it flattens. The air density is by far the highest here, the object is moving fastest here, so most of the energy will be released here and it's closest to the surface, and if enough energy is released it will result in a roughly spherical explosion the same way that a thermonuclear device will even though the bomb itself isn't radially symmetrical. That isn't to say that other parts of the asteroid's trajectory aren't also releasing a shock wave, it's just that only one part of it is .strong enough to flatten trees.