The proof with the first coloring works for this particular problem. The proof for the second doesn’t. That’s really the end of the matter.
You can try to make the argument more formal if that makes more sense to you. Assign a coordinate to each square on the chessboard, so that the top left corner is at $(1,1)$, and the bottom right corner is at $(8,8)$. The “white” squares will now correspond to the squares whose coordinates add to an even number, while the “black” squares will have coordinates that add to an odd number. It’s easy to see that when we remove both previously mentioned corners, we’re left with $32$ black squares, and $30$ white squares. However, a domino, which spans two adjacent squares, must necessarily cover both a white and a black square. Therefore, any valid covering must cover the same number of white and black squares, which means it’s impossible to cover up the whole mutilated board.
Your second coloring not working is just a consequence of the proof above not working for the corresponding modified definitions of the “white” and “black” squares. But a proof not working does not at all imply that another one is false.
For integers $m,n\ge2$ with $mn$ even, let $P(m,n)$ denote the statement: on an $m\times n$ chessboard, any $k$ white squares border at least $k+1$ black squares, provided $0\lt k\lt mn/2$.
The statement $P(m,n)$ implies that an $m\times n$ chessboard will still satisfy Hall's condition (for matching white squares to adjacent black squares) after one black and one white square is removed.
Theorem. If $m,n\in\mathbb N$, $m,n\ge2$, and $mn$ is even, then $P(m,n)$ holds.
The theorem follows directly from the fact that the chessboard graph (whose vertices are the squares, two squares being adjacent if they share a side) has a Hamiltonian circuit. Alternatively, it can be proved by induction using the following lemmas. The easy details are left to the reader.
Lemma 1. $P(2,2)$ and $P(2,3)$ hold.
Lemma 2. Suppose $m,n\in\mathbb N$, $m,n\ge2$, $mn$ even. If $P(m,n)$ holds then $P(n,m)$ holds.
Lemma 3. Suppose $m,n_1,n_2\in\mathbb N$; $m,n_1,n_2\ge2$; and $mn_1$ and $mn_2$ are even. If $P(m,n_1)$ and $P(m,n_2)$ hold, then $P(m,n_1+n_2)$ holds.
Best Answer
Suppose w.l.o.g. that the top left square is black. Divide the board into $2\times 2$ regions. These regions will be three kinds: $A$, $B$, or unmarked. In each $A$ region, we will mark the top left black square. In each $B$ region, we will mark the bottom right black square.
First we assign the regions along the top-right/bottom-left diagonal. These are all $A$ when $n\equiv 2\pmod4$, and the diagonal immediately below it is all $B$. On the other hand, if $n$ is a multiple of $4$, then we make that diagonal all $B$, with the diagonal immediately above it all $A$.
Moving out along parallel diagonals, above each $A$ diagonal is a blank one, followed by another $A$ diagonal. Below each $B$ diagonal is a blank one, followed (if there's still room) by another $B$ diagonal.
To verify that this works, we need to check white squares in the three types regions:
Before I started looking at it diagonally, I couldn't make heads or tails of it. The "row-major" process you describe does appear to generate the same result, although I don't immediately see why it should do that.