Since we know that the block is restricted to move only along the inclined plane, all the force components perpendicular to the inclined plane come in action-reaction pairs( assuming that the inclined plane is stationary and perfectly rigid ) and therefore I shall not include those components in my free body diagrams in order to make things clear.
First, let us take a look at the free body diagram of the block when it is at rest in the absence of the external force F.
In this case, the components perpendicular to the inclined plane ( $mg\cos B$ and N )
cancel each other. Since we are only concerned about the motion along the inclined plane, I have not included them in the diagram. The components along the plane( $mg\sin B$ and $F_s$ ) also come in pairs. The net force on the block is therefore 0 N and the block remains at rest..
*Note: One cannot avoid the case of static friction. If one looks at the region of contact between two surfaces( requires a great extent of magnification ), one would find many ups and downs( mountains ) on each of the surfaces that are locked into each other. This mechanism prevents relative motion. Friction arises as a result of this mechanism. However, at the atomic level, it is the electromagnetic interaction that is responsible for various forces including friction. To assume $\mu_s = 0$ is to assume that the surfaces are smooth. In that case, μk will also be zero.*
Now, let us look at the free body diagram of the block at rest when an external force F is acting on it.
Figure 1 shows the components of the external force. $F\cos B$ acts along the inclined plane and $F\sin B$ acts perpendicular to the plane.
Figure 2 shows various forces acting along the inclined plane. Once again, we are bothered about the motion along the inclined plane and hence ignored the components of force perpendicular to the inclined plane as they always come in action-reaction pairs.
From Figure 2, one can observe that the magnitude of static friction has reduced by $|F\cos B|$.
Note: Static friction is a variable physical quantity.
Even in this case, a net 0 N force is acting on the block and it remains at rest.
Now, as the magnitude of the external force keeps increasing, the magnitude of static frictional force keeps decreasing( provided the block remains at rest ) and the component of the external force up the inclined plane keeps increasing so as to see that the block remains at rest. At one instant, the static frictional force becomes 0 N and $F\cos B = mg\sin B$.
So, as the magnitude of the external force increases, the component of the force up the inclined plane increases.
For $F\cos B > mg\sin B$, a net force $F\cos B - mg\sin B$ acts on the block up the inclined plane. Now, static friction reverses its direction (down the incline) in order to prevent the block from moving up the inclined plane. As $F\cos B$ increases up the inclined plane, static friction increases down the inclined plane in order to prevent relative motion thereby keeping the block at rest as shown below.
At one instant, when static friction reaches its maximum value, an infinitesimal increase in $F \cos B$ will make the block move up the inclined plane and this is when kinetic friction comes into play.
Kinetic friction
If $F\cos B - (mg\sin B + F_s) > F_k$ (kinetic friction), the block accelerates up the inclined plane.
Note: Kinetic friction is a constant physical quantity and it is always less in magnitude when compared to static friction.
Kinetic friction opposes relative motion. So, in this case, the direction of $F_k$ is down the inclined plane( opposite to the direction of motion as shown in figure 3 ).
Analysis can be made by first studying all the forces involved and depending on the direction and magnitude of all the forces involved, one can determine the direction of motion.
Rolling friction is neither static nor kinetic (sliding) friction. It's caused by inelastic forces between the wheels and the road. For example, compressing the tire may take a lot of force, which is pointed up and toward the back of the vehicle. But when the tire expands (as that part of the tire is leaving the road), it may expand with less force, which is directed up and toward the front of the vehicle. Since the rearward component of the compressing regions is less than the forward component of the expanding regions, there is a net rearward force.
A rolling friction coefficient would be much less than either static or kinetic friction. Since you're not given a rolling friction coefficient or a different model for rolling friction, it looks like you can ignore it.
This page (http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html) has some decent diagrams.
Best Answer
To begin with, I don't care much for the diagram you obtained as it implies a gradual transition over time between the maximum static friction friction force and the onset of the constant kinetic friction. In my view a better diagram of the friction plot is one relating friction to applied force as found here: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin
You'll note that the transition between static and kinetic friction is shown as a vertical line, implying a sudden decrease in friction from static to kinetic where the friction force is undefined during the transition. This is consistent with observation. For example, if you try to push a heavy box along on a floor the static friction force prevents movement. When your applied force matches the maximum possible static friction force the box "breaks free" resulting in kinetic friction that providing less resistance than the static friction before breaking free. This is reflected in the friction plot of the link.
As pointed out in the link: "The coefficient (of kinetic friction) is typically less than the coefficient of static friction, reflecting the common experience that it is easier to keep something in motion across a horizontal surface than to start it in motion from rest"
I assume you are talking about the kinetic friction force. The coefficient of kinetic friction is generally presented as being independent of the speed of the sliding object. However, that is generally only true for a range of low speed. As stated in the link "when two surfaces are moving with respect to one another, the frictional resistance is almost constant over a wide range of low speeds"
Bottom line, while we would like to know precisely "the actual mechanism(s)" involved relating friction to surface conditions, speed, and such as you have asked, those mechanisms, to quote the link "defy precise description"
Hope this helps.