I've done some computations and found some relations among the slopes, so I decided to post this as an answer, which can hopefully attract others to explore the relations among the lengths and areas as well.
Suppose $f(x) = ax^4 + bx^3 + cx^2 + dx + e$ with $a \neq 0$ has three local extrema $A, O, B$ and two inflection points $P, Q$. Label these five points $A, P, O, Q, B$ from left to right.
First, let me point out the following trivial facts by virtue of Vieta's formulas: the sum of all the (real and imaginary) roots of $f$ is $-\frac{b}{a}$, the sum of the $x$-coordinates of $A, O$ and $B$ is $-\frac{3b}{4a}$, and the sum of the $x$-coordinates of $P$ and $Q$ is $-\frac{b}{2a}$, so in particular
\begin{equation} \tag{1}
\frac{x_A+x_O+x_B}{x_P+x_Q} = \frac{3}{2}.
\end{equation}
In other words, the average of the $x$-coordinates of the critical points equals the average of the $x$-coordinates of the inflection points.
Since here we are concerned only with slopes (and similarly for lengths and areas), WLOG we can assume $O$ is the origin by performing a translation. A necessary (but not sufficient) condition for this is $f(0) = 0$ and $f'(0) = 0$, so that $d = e = 0$.
Then $$f(x) = ax^4 + bx^3 + cx^2 = x^2(ax^2+bx+c),$$
$$f'(x) = 4ax^3 + 3bx^2 + 2cx = x(4ax^2+3bx+2c),$$
$$f''(x) = 12ax^2 + 6bx + 2c = 2(6ax^2 + 3bx + c).$$
The critical points are
$$O(0, 0),$$
$$A\Bigg(\frac{-3b-\sqrt{9b^2-32ac}}{8a}, \frac{\Big(3b+\sqrt{9b^2-32ac}\Big)^2 \Bigl(-3b^2+16ac-b\sqrt{9b^2-32ac}\Bigl)}{2048a^3}\Bigg),$$
$$B\Bigg(\frac{-3b+\sqrt{9b^2-32ac}}{8a}, \frac{\Big(3b-\sqrt{9b^2-32ac}\Big)^2 \Bigl(-3b^2+16ac+b\sqrt{9b^2-32ac}\Bigl)}{2048a^3}\Bigg),$$
and the inflection points are
$$P\Bigg(\frac{-3b-\sqrt{9b^2-24ac}}{12a}, \frac{\Big(3b+\sqrt{9b^2-24ac}\Big)^2 \Bigl(-3b^2+20ac-b\sqrt{9b^2-24ac}\Bigl)}{3456a^3} \Bigg),$$
$$Q\Bigg(\frac{-3b+\sqrt{9b^2-24ac}}{12a}, \frac{\Big(3b-\sqrt{9b^2-24ac}\Big)^2 \Bigl(-3b^2+20ac+b\sqrt{9b^2-24ac}\Bigl)}{3456a^3} \Bigg).$$
Denote by $m_{XY}$ the slope of the line passing through $X$ and $Y$, and by $m_X$ the slope of the tangent line at $X$. Then
$$m_{AB} = \frac{9b^3-32abc}{64a^2},$$
$$m_{PQ} = \frac{b^3-4abc}{8a^2},$$
which implies that $m_{AB} - m_{PQ} = \frac{b^3}{64a^2}$.
$$m_{OP} + m_{OQ} = \frac{3b^3-14abc}{24a^2},$$
$$m_{OA} + m_{OB} = \frac{9b^3-40abc}{64a^2},$$
$$m_P + m_Q = \frac{b^3-4abc}{4a^2}.$$
Each of the above $5$ quantities is a linear combination of $\frac{b^3}{a^2}$ and $\frac{bc}{a}$, so it is easy to find many ratios that are independent of the coefficients of $f$. The difficulty lies in finding simple and elegant ones such as the following two:
\begin{equation} \tag{2}
m_{PQ} = \frac{1}{2}(m_P + m_Q).
\end{equation}
\begin{equation} \tag{3}
\frac{m_{AB}-m_{OA}-m_{OB}}{m_{PQ}-m_{OP}-m_{OQ}} = \frac{3}{2}.
\end{equation}
I'm looking forward to more insights from this community regarding other invariant quantities.