This result does hold for higher-degree polynomials; you just didn't generalize it correctly. The general statement is as follows. Let $p(x) = (x - a) q(x)$ where $q(x)$ is any differentiable function, and let $r$ be such that $q'(r) = 0$. Then the tangent line at $(r, p(r))$ intersects the $x$-axis at $(a, 0)$.
This is a fairly simple computation. Since $p'(x) = (x - a) q'(x) + q(x)$, it follows that $p'(r) = q(r)$, so the tangent line at $(r, p(r))$ has slope $q(r)$ and hence it intersects the $x$-axis at $(r - \frac{p(r)}{q(r)}, 0) = (a, 0)$. In particular, the above result holds for $q(x)$ a polynomial of any degree greater than or equal to $2$.
(Intuitively, when $q'(r) = 0$, the linear approximation to $p(x)$ at $r$ is $(x - a) q(r)$, so of course it has to hit the $x$-axis at $(a, 0)$. In fact, this way of thinking about it tells you that the $r$ such that $q'(r) = 0$ are the only $r$ for which this occurs.)
In the special case that $q(x)$ is quadratic, $r$ happens to be equal to the average of the roots of $q$. For higher-degree polynomials this is no longer the case. Instead, all you know is that the roots of $q'(x)$ interlace between the roots of $q(x)$.
Some reducible cubics look like
$$(x+r)(x+s)(x+t)$$
and when $r$, $s$ and $t$ are distinct, your method counts them three times.
Best Answer
$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{Frac}$No. For instance, that is impossible (at least, in domains) for polynomials in the form $(x-a)^2(x-b)$ with $a\ne b$.
Added: This is due to the fact that, if such an affine map existed, then your polynomial would have the same number of distinct roots in $\Frac K$ as $x^3+c-l(0)$. However, $x^3+\alpha$ may only have $0$, $1$ or $3$ distinct roots in the field $\Frac K$.