One way to understand how the test works is by looking at the Taylor Series of the function $f(x)$ centered around the critical point, $x = c$:
$$
f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2}(x-c)^2 + \cdots
$$
Note: In your question you said that the n-th derivative is non-zero. Here I'm assuming the n+1-st derivative is the first to be non-zero at $x=c$. It doesn't make a difference, it's just the way I learned it.
If $f'(c) = \cdots = f^{(n)}(c) = 0$ and $f^{(n+1)} \ne 0$, then the Taylor Series ends up looking like this:
$$
f(x) = f(c) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
Consider what happens when you move $f(c)$ to the other side of the equation:
$$
f(x) - f(c) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
What does $f(x) - f(c)$ mean?
- If $f(x) - f(c) = 0$, then $f(x)$ has the same value as it does at $x = c$.
- If $f(x) - f(c) < 0$, then $f(x)$ has a value less than it has at $x = c$.
- If $f(x) - f(c) > 0$, then $f(x)$ has a value greater than it has at $x = c$.
We expect $f(x) - f(c) = 0$ at $x = c$ (the equation reflects this), but we're more interested in what it does on either side of $x = c$. When $x$ is really close to $c$, i.e. $(x-c)$ is a really small number, we can say:
$$
f(x) - f(c) \approx \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1}
$$
because the higher powers of a small number "don't matter" as much.
Concerning local extrema
If $n$ is odd, then our approximation of $f(x) - f(c)$ is an even-power polynomial. That means $f(x)$ has the same behavior - is either less than or greater than $f(c)$ - on both sides of $x = c$. Therefore it's a local extreme. If $f^{(n+1)}(c) > 0$, then $f(x)$ is greater than $f(c)$ on both sides of $x = c$. Otherwise, if $f^{(n+1)}(c) < 0$, then $f(x)$ is less than $f(c)$ on both sides of $x = c$
If, on the other hand, $n$ is even, then our approximation of $f(x) - f(c)$ is an odd-power polynomial centered around $x = c$. Therefore $f(x)$ will be greater than $f(c)$ on one side of $x = c$, and less on the other. That means $x = c$ isn't a local extreme.
Concerning saddle points
Note that if you differentiate both sides of our approximation twice, you get:
$$
f''(x) \approx \frac{f^{(n+1)}(c)}{(n-1)!}(x-c)^{n-1}
$$
If $n$ is even, this is another odd-power polynomial centered around $x = c$. It therefore has opposite behavior on each side of $x = c$, giving you a saddle point.
Following the Taylor development, assuming WLOG the inflection point at $x=0$, we have
$$f(x)\approx f(0)+f'(0)x+f^{(n)}(0)\frac{x^n}{n!}$$
where $n$ is the order of the first nonzero derivative ($n>1$). For $x$ small enough, the higher order terms have no influence. To get an inflection, the last term must be an odd function, so that the curve can cross the straight line $y=f(0)+f'(0)x$.
Below, plots of $1-x, 1-x+x^2,1-x+x^3,1-x+x^4,1-x+x^5$:
Best Answer
One (unsatisfying) answer is that there just aren't that many differential equations we care about with more than 2 derivatives. In the case of ODEs, higher order systems can always be reduced to a 1st order vector ODE, so many systems are only analyzed in their 1st order representations. However, there are notable exceptions (often coming from physics) where 2nd order structure is preferred, such as Newtonian systems, $$y'' = f(y).$$ A natural question then is why are physical systems often 2nd order, which has some interesting discussion here.
If you look at partial differential equations (PDEs), then there are some higher order equations that are very interesting, for example, the KdV equation $$ \frac{\partial u}{\partial t}+\frac{\partial^3 u}{\partial x^3}-6u\frac{\partial u}{\partial x}=0, $$ which appears in the study of shallow water waves, or the biharmonic equation (or Euler beam equation) $$ \frac{\partial^4 u}{\partial x^4} = 0, $$ which appears in solid mechanics when studying some thin surfaces such as plates.
It is worth noting that these higher order derivatives aren't "inherently physical," by which I mean they often come up by combining physical laws (often 2nd order) with some assumption on material behavior that then allows one to close the system and possibly end up with a system with higher derivatives. For an example, when deriving the equations of solid mechanics for a thin plate you end up with a term like $\frac{\partial^2 M}{\partial x^2}$, where $M$ is a bending moment. This relation comes from careful analysis of the system and application of Newton's Laws. One can then close the system by making some assumptions on the material properties, such as linear elasticity, to derive the relation $M\propto\frac{\partial^2 u}{\partial x^2}$. One can then combine both relations to obtain a 4th order system, even though no physical law explicitly stated a 4th order relationship between any of the quantities involved.