Linearity (or lack thereof) of the Poynting vector Maxwell's equations are linear. If we have a solution for the electromagnetic fields $\vec{E},\vec{H}$, and another solution $\vec{E}',\vec{H}'$, then $\vec{E}+\vec{E}',\vec{H}+\vec{H}'$ is also a solution.
The Poynting vector is nonlinear. I see no mathematical reason to expect that
$$\vec{E}\times \vec{H} + \vec{E}'\times \vec{H}' = (\vec{E}+\vec{E}')\times (\vec{H}+\vec{H}')$$
But this seems ... surprising. Do the cross terms $\vec{E}\times \vec{H}'$ and $\vec{E}'\times \vec{H}$ cancel out, somehow? Or is it really true that we learn nothing about the power flow of $\vec{E}+\vec{E}',\vec{H}+\vec{H}'$ just by knowing the power flows in $\vec{E},\vec{H}$ and $\vec{E}',\vec{H}'$?
 A: Consider Young's double slit experiment? Each of the slits is effectively a "source" for the EM field between the slits and the screen. And the pattern we see on the screen represents the power delivered by the combined sources at each point on the screen.
Now consider the possibility of introducing an optical flat in the path to one of the slits, so by changing the thickness of the flat, we can change the phase of the wave reaching that slit relative to the other. As we adjust the flat thickness, we'll see the peaks of the pattern on the screen shift their positions.
So, without knowing the phase of $\vec{E}$ from a source, we can't know how it will interfere with the $\vec{E}$ from another source. 
And just knowing the power due to each source individually does not tell us how much power will be delivered to a certain surface when the two sources are combined.
A: We learn something (for example, a limit on magnitude that the Poynting vector of the sum field cannot surpass), but not all details ( components of the Poynting vector for the sum field). Poynting vectors of two different fields do not determine Poynting vector of the sum field. Mathematically, products $a_1b_1, a_2b_2$ do not determine product $(a_1+a_2)(b_1+b_2)$.
Equation that $(E,H)$ obeys is linear, so sum of solutions for $(E,H)$ is a solution of that equation too. Equation that $\vec{E}\times \vec{H}$ obeys is not linear, so sum of solutions for $\vec{E}\times \vec{H}$ is not, in general, a solution.
A: The mistake is that you omitted the terms $ \vec{E}\times \vec{H}'$ and + $\vec{E}'\times \vec{H}$ in the left hand side. 
