Calculating power measurement error from V and I error margins
Oh yes, you're definitely missing something. You can't focus on the errors without looking at the "non-errors". Let's use your second example, where you got a power error of 1.404 watts. What is the "non-error" component?

Just for grins, assume that the real voltage was 100 volts and the real current 20 amps. Then the real power was 2,000 W, but the inaccurate readings were 101.17 volts and 21.2 A, for an incorrect power calculation of 2144.8 watts, or a power error of 144.8 watts. 

If you consider each measurement as a base plus an error, and the errors are normalized with respect to the base (such as you would do with percentage errors) then $$(1 + x)(1 + y) = 1 + x + y + xy$$ and the error term is the sum of the errors plus their product. For small errors the product term is negligible, and the error is simply the sum of the errors.

For non-normalized errors, as in your example, you need to compute $$P = (V + {\delta}v)(i+{\delta}i) = Vi + V{\delta}i + i{\delta}V +{\delta}V{\delta}i $$ and the power error E is simply $$E = V{\delta}i + i{\delta}V +{\delta}V{\delta}i $$

So, yes, your reservations were correct.



Figure 1. A graphical representation of the \$VI\$ term (green), \$V \delta i \$ term (red), \$ i \delta V \$, term (blue) and \$ \delta V \delta i \$ term (white). It can be seen that, in this case, the \$ \delta V \delta i \$  term contributes very little to the overall error and that about 80% of the error is due to the \$V \delta i \$ term.