The problem with this question is that static friction and kinetic friction are not fundamental forces in any way-- they're purely phenomenological names used to explain observed behavior. "Static friction" is a term we use to describe the observed fact that it usually takes more force to set an object into motion than it takes to keep it moving once you've got it started.
So, with that in mind, ask yourself how you could measure the relative sizes of static and kinetic friction. If the coefficient of static friction is greater than the coefficient of kinetic friction, this is an easy thing to do: once you overcome the static friction, the frictional force drops. So, you pull on an object with a force sensor, and measure the maximum force required before it gets moving, then once it's in motion, the frictional force decreases, and you measure how much force you need to apply to maintain a constant velocity.
What would it mean to have kinetic friction be greater than static friction? Well, it would mean that the force required to keep an object in motion would be greater than the force required to start it in motion. Which would require the force to go up at the instant the object started moving. But that doesn't make any sense, experimentally-- what you would see in that case is just that the force would increase up to the level required to keep the object in motion, as if the coefficients of static and kinetic friction were exactly equal.
So, common sense tells us that the coefficient of static friction can never be less than the coefficient of kinetic friction. Having greater kinetic than static friction just doesn't make any sense in terms of the phenomena being described.
(As an aside, the static/kinetic coefficient model is actually pretty lousy. It works as a way to set up problems forcing students to deal with the vector nature of forces, and allows some simple qualitative explanations of observed phenomena, but if you have ever tried to devise a lab doing quantitative measurements of friction, it's a mess.)
You're very close. The vector giving the friction force has magnitude $\mu_k F_N$ and is opposite the direction of travel, so it can be written as the product of $\mu_k F_N$ with a unit vector pointing in the direction opposite the direction of travel. Since the velocity $\vec v$ is in the direction of travel, the unit vector $-\vec v/v$, where $v$ is the object's speed, points opposite to the direction of travel, so the force of friction can be written as
\begin{align}
\vec f_k = -\mu_k F_N \frac{\vec v}{v}
\end{align}
The issue with your last expression is that $\vec F_N\cdot \hat d = 0$ since the normal force is perpendicular to the surface, and the direction of travel is parallel to the surface.
Best Answer
The normal force does decrease with angle. This does not mean that the coefficient of friction changes:
We can, depending on the angle $\theta$ of the slope, split the gravitational force $F_g = mg$ acting upon a thing with mass $m$ resting on the slope into the normal force $F_n = mg \cos(\theta)$ and the force pointing down the slope, $F_s = mg\sin(\theta)$.
Now, the coefficient of friction is a property of materials, and does not change with the angle - but it is the case that the friction force will decrease since it is $F_k = \mu_kF_n$. The "greater propensity" of things to slide down steeper inclined slopes is due to the friction force decreasing, and due to the force pointing down the slope increasing with increasing angle.