When can we add multiple reactions to compute yield of products in an equilibrium?
When you combine the reactions and the equilibrium constants like you do, you're calculating the amount of reaction which goes all the way to completion (the fraction which reacts in both steps).

The complication that you're seeing is that the protons are not just coming from the $\ce{H_2A}$ which completely dissociates, but also a (rather substantial) portion of the molecule which undergoes only partial dissociation.

To see this, think about what happens when the hydrogens in the molecule are distinguishable: for example, the dissociation of $\ce{HDA}$ (hemi-deuterated compound), where the reactions are 

$$\ce{HDA &lt;=&gt; DA^{-} + H^{+}} $$
$$\ce{DA^{-} &lt;=&gt; A^{2-} + D^{+}} $$

(in reality this wouldn't be the obligate order, but just assume it is for a thought experiment).

The $10^{-15}$ equilibrium constant is only applicable for the formation of $\ce{D^{+}}$ - you still have the rest of the $\ce{H^{+}}$ from the first step, which will also contribute to the pH.

So your assumption in your last step that $\ce{[H^{+}] = 2 [A^{2-}]}$ is incorrect, thus resulting in an incorrect result.


Further explanation

Let's actually work through the $\ce{HDA}$ case, assuming (incorrectly) that the dissociation has an obligate order. You start off with a certain amount of $\ce{HDA}$ ($T = 0.1M$). Now, imagine that at equilibrium we get $x$ total turnovers for the first reaction. A certain amount of $\ce{DA^-}$ (call it $y$) is then going to go and dissociate further. So in the end at equilibrium, we have the following concentrations:

$$\ce{[HDA]} = T-x$$
$$\ce{[H^+]} = x$$
$$\ce{[DA^-]} = x-y$$
$$\ce{[D^+]} = y$$
$$\ce{[A^{2-}]} = y$$

Now we can set up our equilibrium equations:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} = \frac{(x)(x-y)}{T-x} = 10^{-5} M$$
$$\frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HD^{-}}]} = \frac{(y)(y)}{x-y}= 10^{-10} M$$

We have a system of two equations with two unknowns. We can thus solve for $x$ and $y$ and get $x=0.995\cdot10^{-3} M$ and $y = 3.15\cdot10^{-7} M$.

Now, let's do the combined equation. First off, we can see where it comes from:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} \cdot \frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{DA^{-}}]} = \frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = 10^{-5} M \cdot 10^{-10} M = 10^{-15} M^2$$

So we set up the equation:

$$\frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = \frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

Um ... we have two unknowns, but only a single equation. We can't solve this without additional information. We can't assume that $[\ce{H^{+}}] = [\ce{D^{+}}]$, or that all the $\ce{HDA}$ which disappeared went completely to $\ce{A^{2-}}$ - some of it can (has to) form $\ce{DA^-}$ which doesn't react further. The equation is valid (if you substitute $x=0.995\cdot10^{-3} M$ into the equation, as determined above, you'll get the correct answer for $y$), but only if we have some way of knowing how much of the $\ce{HDA}$ reacted and didn't go completely to completion. 

The picture is further complicated because we're not dealing with distinguishable $\ce{H^+}$ and $\ce{D^+}$ which dissociate in an obligate order, but a single species which is involved in both reactions. That is, instead of 

$$\frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

we would have 

$$\frac{(x+y)(x+y)(y)}{(T-x)} = 10^{-15} M^2$$

Again, the combined equilibrium constant is perfectly correct, and accurately expresses the relationship between the concentrations at equilibrium. It's just that the additional assumptions about the relationships between the concentrations that you're making are not. Specifically, $[\ce{H^{+}}] \neq 2 \cdot [\ce{A^{2-}}]$. Note this isn't particularly specific to this sort of combined equilibrium. Any additional reaction you're not considering can change the concentrations and make your assumptions less than correct - not just formation of intermediates which don't go all the way to completion to side reactions that draw off substrates, reactants or products into other forms.

The equilibrium constant tells you the relationship between the species concentrations, but the concentrations as they actually are in solution, not what the reaction generates. The latter is only true when the reaction in question is the only one under consideration.