Is there any way to prove this assumption?

Here is a configuration satisfying the conditions of the question

$$ \begin{array} { c c c } ak & k & jk & j & ij\\ a & x & x & x & i\\ ab & de & e &ef & hi\\ b & dy & y & fy & h\\ bc & cd &cg & fg & gh\\ \end{array} $$

Notice that the triminos from $a$ to $k$ give us a chain of 11 triminos where each one overlaps with the 2 adjacent ones ($k$ and $a$ are adjacent), hence each adjacent pair of letters have to be on distinct layers.
Thus, it's not possible to split these 11 triminos onto 2 distinct layers. (An odd-chain is not a bipartite graph.)

Hence, it is true that "Not all arrangements are composed of 2 disjoint layers".

Original writeup:

Caveat: I'm not satisfying the condition of "each square is covered by 1 or 2 triminos". I've not investigated if there's a way to modify this example easily. (EG In this example, the bottom left square cannot be covered)

Note: With the relaxation of this condition, we can place 16 triminos (without covering the center square). Hence, this is an important condition to consider, which suggests the solution as listed in the document.

Here is a $5 \times 5 $ board where each square is covered by at most 2 triminos.

We have a chain of 9 triminos ($a, b, c, \ldots i$), where each trimino covers the adjacent 2 triminos (and $i$ covers $a$ too).
Hence we cannot split them up into 2 disjoint layers.
(No way to split an odd-chain into a bipartite graph.)

$$ \begin{array} { c c c } a & & & & \\ ai & i & hi & h & gh\\ ab & bc &be & &g \\ &c & ef & f& fg\\ &cd &de &d & \\ \end{array} $$