F - measure derivation (harmonic mean of precision and recall)
why the equality of both partial derivatives correspond to these
  hypothesis. I would rather understand when one partial derivative
  equals the other partial derivative multiplied by minus one.


Your intuition of "trading off" by "subtracting" a value is correct when you speak in terms of $\Delta R$ and $\Delta P$ (as you yourself noticed in the edit), but is not valid if you speak in terms of derivatives.

Think of it this way: a partial derivative with respect to a parameter tells you by how much your metric changes if you improve that variable. By the very design, $F$ will always increase if you improve either $R$ or $P$ - its partial derivatives are positive everywhere.

They are not equal everywhere, however. At some configurations $(R,P)$ you would gain more by improving one parameter rather than the other. Say, if your $R$ is much lower than $P$, your $F$-score would increase more if you improved the $R$ a bit rather than $P$.

In the case of an $F_1$ metric, $R$ and $P$ enter the formula symmetrically and it is kind-of easy to see that whenever $R=P$, either of the components contributes equally to the increase of $F$. In fact, the set of points where $R$ and $P$ are "equal in importance" is exactly the line $R=P$ (which can also be written as $R/P=1$, if you prefer).

Now consider the equation for $F_\beta$:

$$\frac{\text{constant}}{\beta^2\frac{1}{R} + \frac{1}{P}}.$$

Let us see how much it changes in response to increasing $R$ or $P$. As both parameters are in the denominator, the change in value of $F_\beta$ is proportional to the change in the value of its denominator, hence we may simplify a bit and only consider the change in

$$\beta^2\frac{1}{R} + \frac{1}{P}$$

You can now compute that a $dx$ increase in $R$ would decrease the value of the expression above by $\beta^2\frac{dx}{R^2}$. A similar increase in $P$ would decrease the value of the above expression by $\frac{dx}{P^2}$. There is a set of points $(R,P)$ where these effects are exactly equal. These are the points where 
$$\beta^2\frac{dx}{R^2} = \frac{dx}{P^2} \quad\Rightarrow\quad \frac{R}{P}=\beta$$

When $\beta &gt; 1$ you are effectively putting more marginal importance on recall, in the sense that until your recall is at least $\beta$ times the precision, it is always beneficial to work on it, rather than improving precision.


  What am I missing?


You are missing the fact that the condition $\frac{\partial F_\alpha}{\partial P} = \frac{\partial F_\alpha}{\partial R}$ does not hold everywhere, but just on the set of points $R=\beta P = \sqrt{\frac{1-\alpha}{\alpha}}P$. This line intersects the level curve at just a single point.

The line with slope -1 would be the tangent to the level line at that point - the green dashed line on the illustration below.