(Updated question) Problem with Fold-change mean compared to absolute data (in qPCR)
Sample 1. 4096 / 1024 = 4x more B
Sample 2. 16384 / 1024 = 16x more B
The average amount more B = (16 + 4) / 2 = 10x more B

That is not the right way to go about. You are calculating average expression of B in both samples. It does not mean there is 10x more B. There is just 4x more B in Sample-2 relative to Sample-1.
Gene-A does not change its expression between samples and can be used as a reference gene.
If you calculate by threshold cycle the fold change $ = 2^{8-6} = 4x$
See this post too.
The average is calculated between replicates, not between different experimental conditions.
Sometimes the average of two experimental samples is calculated — this is typically done in a MA-plot analysis which is done to filter out genes which show very high fold changes just because of their overall low expression (1:4 vs 100:300).
Answer to your Edit
You should not take the average (arithmetic mean) of Ct. They don't scale linearly with expression (and so is the converse).
For any non-linear function $f(x)$:
$$f(x+y) \ne f(x)+f(y)$$
$$f\left(\frac{x+y}{2}\right) \ne \frac{f(x)+f(y)}{2}$$
Furthermore, for a convex function:
$$f(E[x]) \le  E[f(x)] $$
$$ Jensen's\ inequality $$
where $E[x]$ is expected value of $x$
$a^x$ is a convex function with respect to $x$ ($a$ is any real number &gt; 1). So you can apply Jensen's inequality to Ct values and expression.