The information you get is when you only have $f''(x)=0$ then you now that $f$ is a linear function, because this is an "if and only if"-relation.
That is: If $f(x)$ is linear, then $f''(x)=0$.
And: If $f''(x)=0$, then $f(x)$ is linear.
That reads: $f(x)$ is linear $\leftrightarrow$ $f''(x)=0$
An asymptote (horizontal or vertical) occurs when a line fits the curve at infinity.
$$\lim_{x\to\infty}(f(x)-(ax+b))=0.$$
The angular coefficient can be found by writing
$$\lim_{x\to\infty}\frac{f(x)-(ax+b)}x=0,$$
or
$$\lim_{x\to\infty}\frac{f(x)}x=a,$$ if that limit exists, and the intercept from
$$\lim_{x\to\infty}(f(x)-ax)=b,$$ if that limit exists.
The first limit can also be evaluated by the L'Hospital rule (provided its conditions of application are fulfilled):
$$a=\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}\frac{f'(x)}1.$$
This is how the first derivative enters into play. For this reason, one also says that an asymptote is a tangent at infinity. In other words, there is an asymptote if the tangent to the curve tends to a particular straight line for $x$ going to infinity.
$a$ exists if the function doesn't grow faster than $x$, i.e. if the slope of the function is bounded and convergent (to zero in the horizontal case). This is a necessary but not sufficient condition. Then $b$ exists if the difference between the function and the approximation $f(x)-ax$ is itself bounded and convergent.
A few examples:
$f(x)=x^2$ has no asymptote because $\dfrac{x^2}x$ is unbounded (so is $f'(x)=2x$).
$f(x)=x\sin(x)$ has no asymptote because $\dfrac{x\sin x}x$ does not converge (nor does $\sin x+x\cos x$).
$f(x)=\sqrt x$ has no asymptote, because $a=0$ is true, but $b$ does not exist. In fact, $\lim_{x\to\infty}(\sqrt x-0x)=\infty$, and one can say that the asymptote is a line "at infinity".
$f(x)=\sqrt{x^2+1}$ has an asymptote because $a=1$ holds (both from $\dfrac{f(x)}x$ and $f'(x)$), and $\lim_{x\to\infty}\left(\sqrt{x^2+1}-x\right)=\dfrac12$.
Notice that in the above discussion, the higher order derivatives are never used. In particular, their respective signs are playing no role. For example, $\dfrac{\sin x}x$ has an horizontal asymptote, while all its derivatives are alternating.
So the answer to your title question is "they don't".
Best Answer
In a very natural sense, you can! If $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = L$ is some real number, then it makes sense to define $f(\infty) = L$, where we identify $\infty$ and $-\infty$ in something called the one-point compactification of the real numbers (making it look like a circle).
In that case, $f'(\infty)$ can be defined as $$f'(\infty) = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$ When you learn something about analytic functions and Taylor series, it will be helpful to notice that this is the same as differentiating $f(1/x)$ at zero.
Notice that this is actually not the same as $\lim_{x \to \infty} f'(x)$.
These ideas actually show up quite a bit in analytic capacity, so this is a rather nice idea to have.
I wanted to expand this answer a bit to give some explanation about why this is the "correct" generalization of differentiation at infinity. and hopefully address some points raised in the comments.
Although $\lim_{x \to \infty} f'(x)$ might feel like the natural object to study, it is quite badly behaved. There are functions which decay very quickly to zero and have horizontal asymptotes, but where $f'$ is unbounded as we tend to infinity; consider something like $\sin(x^a) / x^b$ for various $a, b$. Furthermore, $\lim_{x \to \infty} f'(x) = 0$ is not sufficient to guarantee a horizontal asymptote, as $\sqrt{x}$ shows.
So why should we consider the definition I proposed above? Consider the natural change of variables interchanging zero and infinity*, swapping $x$ and $1/x$. Then if $g(x) := f(1/x)$ we have the relationship
$$\lim_{x \to 0} \frac{g(x) - g(0)}{x} = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$
That is to say, $g'(0) = f'(\infty)$. Now via this change of variables, neighborhoods of zero for $g$ correspond to neighborhoods of $\infty$ for $f$. So if we think of the derivative as a measure of local variation, we now have something that actually plays the correct role.
Finally, we can see from this that this definition of $f'(\infty)$ gives the coefficient $a_1$ in the Laurent series $\sum_{i \ge 0} a_i x^{-i}$ of $f$. Again, this corresponds to our idea of what the derivative really is.
* This is one of the reasons why I used the one-point compactification above. Otherwise, everything that follows must be a one-sided limit or a one-sided derivative.