I believe I've found a solution that is quite basic and should be approachable by people with an elementary understanding of mathematics. Color each square of the grid with colors $1$ through $k$, as follows. In the image below, $k=7$.
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c}
\hline
1&2&3&4&5&6&7&1&2&3&\dots\\\hline
2&3&4&5&6&7&1&2&3&4&\dots\\\hline
3&4&5&6&7&1&2&3&4&5&\dots\\\hline
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\
\end{array}$$
Then each tile will contain exactly one color of each kind. It follows that
If the grid can be tiled by $k\times 1$ rectangles, then it contains the same number of squares of each color.
Let's agree that the grid is $m$ rows by $n$ columns.
If $m$ is divisible by $k$, say $m=ck$, then for each color and each column there will be exactly $c$ squares of that color in that column. Hence, the total number of squares of each color is the same. Something similar holds when $n$ is divisible by $k$.
Now, suppose neither $m$ nor $n$ is divisible by $k$, and write $m=ck+r$ and $n=dk+q$, with $1\leq r,q\leq k-1$. The subgrid $ck\times n$ contains an equal number of squares of each color, by the previous argument. Consider then the bottommost strip, of size $r\times n$.
Once again, the subgrid of size $r\times dk$ contains an equal number of squares of each color. We are finally left with a subgrid of size $r\times q$.
Assume without loss of generality that $r\leq q$. Renumbering if needed, it should look something like this:
$$\begin{array}{|c|c|c|c|}
\hline
1&2&\dots&q\\\hline
2&3&\dots&q+1\\\hline
\vdots&\vdots&\ddots&\vdots\\\hline
r&r+1&\dots&r+q-1\\\hline
\end{array}$$
Since $r\leq q$, it's easy to see that color $q$ shows up in exactly $r$ squares -- once on each row (consider the antidiagonal). On the other hand, color $1$ definitely does not show up on the second row, so it can appear at most $r-1$ times.
This shows that, as a whole, the $m\times n$ grid does not contain the same number of squares of each color. Therefore, if $k$ divides neither $m$ nor $n$, the $m\times n$ grid cannot be tiled by $k\times 1$ rectangles.
In an idealized $n\times m$ room using $1\times 1$ tiles it would take $0$ cuts to lay tiles rectilinear pattern.
In an idealized $\sqrt2n \times \sqrt 2 m$ room, it will take $n+m$ cuts to lay tiles on the 45.
However rooms are never ideal. With the rectilinear pattern, it is only necessary to cut tiles along two walls
With the 45 pattern you must cut tiles along 4 walls.
When you can cut tiles exactly in half then you can tile that wall with fewer tiles as each diagonal is $\sqrt 2$ units long. But on the opposite wall, you need twice as many tiles, and Each tile piece is on average $\frac {\sqrt2}{2}$ units long.
Many of the trimmings are not reusable. Looking at your pattern when you cut with one big piece and one small piece, you keep using the big piece of one colored tile and the small piece of the other colored tile.
If the tiles are uniform in coloration, you can reuse more of your cuttings.
Putting this together, double looks to be more than just a rule of thumb. It would take some interesting coincidences to be too far off from that.
Best Answer
Recalling Kenyon's work on this, I found these comments in an old MO post