The first two proofs only work for a fully populated heap of height h, which contains n=2h-1 items. The third proof also works for partial heaps. Throughout this answer, i will use two functions called BuildCost and MergeCost. The build cost is the cost for constructing a heap of height h from scratch. The merge cost is for merging two heaps of height h and one unsorted item into a single heap of height h+1.
Your approach, revisited
I used h-1 and n+1 instead of floor/ceiling, so my proof is only valid for a fully populated heap as described in the intro. All divisions and logarithms are exact in my solution.
$$
\begin{align*}
\text{BuildCost}(h) &= \sum_{i=0}^{h-1}\frac{2^{h-1}}{2^{i+1}}\cdot\text{MergeCost}(i) \\
&= \sum_{i=0}^{h-1}\frac{2^{h-1}}{2^{i+1}}\cdot2i \\
&= 2^h\sum_{i=0}^{h-1}\frac{i}{2^i} \\
&= (n+1)\sum_{i=0}^{h-1}\frac{i}{2^i}
\end{align*}
$$
At this point, we can use the identity:
$$\sum_{m=0}^M\frac{m}{2^m} = 2-\frac{M+2}{2^M}$$
to solve the build cost as:
$$
\begin{align*}
\text{BuildCost}(h) &= 2^h\sum_{i=0}^{h-1}\frac{i}{2^i} \\
&= 2^h\left(2 - \frac{h-1+2}{2^{h-1}} \right)\\
&= 2\cdot2^h - 2h -2 \\
&= 2(n+1) - 2h - 2 \\
&= 2n - 2h \\
\end{align*}
$$
Alternative: proof by induction
Here is a proof by induction, which confirms the results above. I like it better because it doesn't rely on a non-obvious identity like the other proof.
Induction hypothesis:
$$
\begin{align*}
\text{BuildCost}(h) &= 2n - 2h\\
\text{MergeCost}(h) &= 2h \\
n &= 2^h-1
\end{align*}
$$
The base case is a heap of height one, which requires zero comparisons to build:
$$
\text{BuildCost}(1) = 2(2^1-1) - 2\cdot1 = 0
$$
The induction step is:
$$
\begin{align*}
\text{BuildCost}(h+1) &= 2(2^{h+1}-1) - 2(h+1)\\
&= 4\cdot2^h -2h -4 \\
&= 4\cdot2^h -4h -4 &+& 2h \\
&= 2(2\cdot2^h -2h -2) &+& 2h \\
&= 2\cdot(2(2^h-1)-2h) &+& 2h \\
&= 2\cdot\text{BuildCost}(h) &+& \text{MergeCost}(h)
\end{align*}
$$
Extension to partial heaps
The build cost for a full heap is 2n-2h. We can think of the -2h term as a "merge budget" so that when we merge the heap, we're still below the 2n limit. If we merge two heaps of the same height, both sides bring a budget of 2h and the unsorted item brings a budget of 2, so we end up with 4h+2 of which the merge itself consumes 2h. What remains is 2(h+1) and this is exactly the merge budget that we want to pass up the tree. So far, this paragraph is a repeat of the induction step.
Now let's do the induction step again, but this time the left child has height h and the right child has height h-1. The left child has nl items (fully populated or not) and build cost of 2nl-2h. The right child has nr items (fully populated or not) and build cost of 2nr-2(h-1). My working hypothesis is that all heaps of height h have a build cost of 2n-2h regardless if fully populated or not. This is false, but for now let's pretend it's true.
$$
\begin{align*}
\text{BuildCost} &= \text{BuildCost}(h) + \text{BuildCost}(h-1) + \text{MergeCost}(h)\\
&= 2n_l - 2h + 2n_r - 2(h-1) + 2h \\
&= 2n_l + 2n_r + 2 - 2h \\
&= 2(n_l + n_r + 1) - 2h \\
&\neq 2(n_l + n_r + 1) - 2(h+1)
\end{align*}
$$
The merged heap has nl+nr+1 items and height h+1. According to my working hypothesis, it should have a build cost of 2(nl+nr+1)-2(h+1). This is wrong, the actual build cost requires two extra comparisons. Using the terminology of "merge budget" again, we can clearly see what went wrong: The shorter child's merge budget was only 2(h-1), yet the merge required 2h comparisons in the worst case (recursion into the taller child) and we're now two comparisons short of a balanced budget. In conclusion, a heap of n items has a build cost of:
$$\text{BuildCost} = 2n-2h+2k $$
where h is the height and k is the number of asymmetric mergers (left child one taller than right child) that occurred during construction of the heap. Now we have to prove that k<h or, in other words, that there is at most one asymmetric merger per (non-leaf) layer. This is indeed true. Here is a small example of a heap with n=10, h=4. The letter x denotes a missing leaf (compared to a fully populated heap).
1
2 3
4 5 6 7
8 9 10 x x x x x
Since we populate the heap breadth first and from left to right, there can be only one partially populated node per layer. In this example, only nodes 1 and 5 have children of unequal height, so k=2. Here's another example where nodes 1, 2, and 4 have children of unequal height, so k=3. That's the maximum asymmetry we can achieve in a heap of height 4.
1
2 3
4 5 6 7
8 x x x x x x x
But wait, there's more. We can find the asymmetric mergers from the binary representation of the number of missing leafs. In the n=10 example, we had 5 missing leafs. 5 is 101 in binary, which tells us there's an asymmetric merger in the top and bottom layer but not in the middle layer (we ignore the leaf layer here, so top/middle/bottom refers to the non-leaf layers). The n=8 example had 7 missing leafs, which is 111 in binary. I think you can see a pattern. Finally, the build cost of an arbitrary heap is:
$$\text{BuildCost} = 2n-2h+2\cdot\text{PopCount}(2^h-1-n) \le 2n-2 $$
