To expand on my comment, your equation holds only if $A$ and $B$ are uncorrelated. More generally, if $F$ is a function of uncorrelated variables $A$ and $B$, then
$$(\delta F)^2 = \left(\frac{\partial F}{\partial A}\delta A\right)^2 + \left(\frac{\partial F}{\partial B} \delta B\right)^2$$
If you plug in $F=AB$, then you find that
$$ \left(\delta F\right)^2 = (B\cdot \delta A)^2 + (A \cdot \delta B)^2$$
or
$$ \left(\frac{\delta F}{F}\right)^2 = \left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2 $$
In your case, $A$ and $B$ are the same, so they are the exact opposite of uncorrelated. In your case, the appropriate thing to do would be to consider $F$ to be a function of a single variable, and simply let
$$\delta F = \left|\frac{\partial F}{\partial A}\right| \delta A$$
$$\frac{\delta F}{F} = \left|\frac{1}{F}\frac{\partial F}{\partial A}\right| \delta A$$
To answer the question at a deeper level, we model experimentally measured quantities as continuous random variables, with our experimental uncertainties corresponding to their standard deviations.
Let $X$ be a random variable with expected value $\mathbb{E}[X]=\mu_X$ and variance $Var(X)\equiv \mathbb{E}\big[ (X-\mu_X)^2\big]=\sigma_X^2$, and let $g$ be a function of $X$. We can expand $g$ in a Taylor series around $\mu_X$:
$$g(X) = g(\mu_x) + g'(\mu_X)\cdot (X-\mu_X) + \frac{1}{2}g''(\mu_X) (X-\mu_X)^2 + \ldots $$
Truncating after the linear term, we write
$$g(x) = g(\mu_X) + g'(\mu_X)\cdot (X-\mu_X)$$
We can now calculate the mean and variance of $g$:
$$\mathbb{E}[g(X)] = g(\mu_X) + g'(\mu_X)\mathbb{E}[X-\mu_X] = g(\mu_X) + g'(\mu_X)\cdot (\mu_X-\mu_X) = g(\mu_X)$$
$$Var(g(X)) = \mathbb{E}\big[\big(g(X)-g(\mu_X)\big)^2\big] = \big(g'(\mu_X)\big)^2 \cdot \mathbb{E}\big[(X-\mu_X)^2\big] = \left(\frac{dg}{dX}\cdot \sigma_X\right)^2$$
This is where the single-variable error propagation formula comes from. But now, consider a function $F$ of two variables $X$ and $Y$, with respective means $\mu_X,\mu_Y$ and variances $\sigma_X^2,\sigma_Y^2$, and covariance
$$Cov(X,Y) = \mathbb{E}\big[(X-\mu_X)\cdot(Y-\mu_Y)\big]$$
We Taylor expand $F$ to linear order:
$$F(X,Y) = F(\mu_X,\mu_Y) + \frac{\partial F}{\partial X}(X-\mu_X) + \frac{\partial F}{\partial Y} (Y-\mu_Y) $$
Repeating the earlier steps, the mean of $F$ is
$$\mathbb{E}\big[F(X,Y)\big] = F(\mu_X,\mu_Y) $$
The variance, however, develops a slight subtlety. Notice that
$$\left(F(X,Y)-F(\mu_X,\mu_Y)\right)^2 = \left(\frac{\partial F}{\partial X}\right)^2(X-\mu_X)^2 + \left(\frac{\partial F}{\partial Y}\right)^2(Y-\mu_Y)^2 + 2\frac{\partial F}{\partial X}\frac{\partial F}{\partial Y}(X-\mu_X)(Y-\mu_Y)$$
It follows that
$$Var\big(F(X,Y)\big) = \left(\frac{\partial F}{\partial X} \sigma_X\right)^2 + \left(\frac{\partial F}{\partial Y}\sigma_Y\right)^2 + 2\frac{\partial F}{\partial X} \frac{\partial F}{\partial Y} Cov(X,Y)$$
If $X$ and $Y$ are uncorrelated, then $Cov(X,Y)=0$, and so we get our nice simple formula again. However, if $Cov(X,Y)\neq 0$, we need to take it into account.