How can one resolve an apparent paradox regarding the uncertainty of the product of two measured quantities? Suppose one has three quantities $X$, $X_1$ and $X_2$, such that $X = X_1X_2$. Since percentages uncertainties of products just add up we have:
$$\frac{\delta X}{X} = \frac{\delta X_1}{X_1} + \frac{\delta X_2}{X_2}$$
therefore:
$$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} - \frac{\delta X_2}{X_2}$$
However, since $X_1 = \frac{X}{X_2}$, and percentage uncertainties of ratios also add up:
$$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} + \frac{\delta X_2}{X_2}$$
Which directly contradicts the formula given above. How can one resolve this paradox?
Here's what I have thought about to try to resolve the above paradox so far:
Let $X, X_1, X_2$ be random variables such that $X = X_1X_2$, if the first and second moments of $X, X_1$ and $X_2$ exist, then:
$$\frac{Var[X]}{(E[X_1]E[X_2])^2} = \frac{Var[X_1]}{E[X_1]^2} + \frac{Var[X_2]}{E[X_2]^2} + \frac{Var[X_1]Var[X_2]}{(E[X_1]E[X_2])^2}$$
In practice the third term is negligibly small, also errors in experiments are usually uncorrelated so we have:
$$\frac{Var[X]}{E[X]^2} \approx \frac{Var[X_1]}{E[X_1]^2} + \frac{Var[X_2]}{E[X_2]^2}$$
However, it seems to me that the formula for the ratio of two random variables has to be a lot more complicated in general, so I couldn't derive it. So that's where I got stuck.
 A: 
However, since $X_1 = \frac{X}{X_2}$, and percentage uncertainties of
  ratios also add up: $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} + \frac{\delta X_2}{X_2}$$

Where did you get this formula?
Even for multiplication you have to understand that the "error propagation" equations are not exact. They are approximate, order of magnitude, back of an envelope kind of relations. You may look at "propagation of uncertainties" subject in Google to see how they come up with these relations. In any case, your statement about the propagation of errors in division is unusual at best.
Applying the taylor expansion approach from the second line, you can get easily 
 $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} - \frac{\delta X_2}{X_2}$$, which is compatible with your base case of $X_1X_2$. This is how it's done.
$$\left(\frac{x}{y}\right)'=\frac{x'}{y}-\frac{xy'}{y^2}$$
Divide both sides by $\frac{x}{y}$ and get:
$$\frac{\left(\frac{x}{y}\right)'}{\frac{x}{y}} = \frac{x'}{y}\frac{y}{x}-\frac{xy'}{y^2}\frac{y}{x}=\frac{x'}{x}-\frac{y'}{y}$$
If you call $z=\frac{x}{y}$ then you get
$$\frac{z'}{z}=\frac{x'}{x}-\frac{y'}{y}$$. Q.E.D.
