Matrix decomposition
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Adapted from Chapter 1 of “Boolean function complexity” by Stasys Jukna.
A boolean matrix $M$ is primitive if it’s the indicator of a combinatorial rectangle: $M_{ij} = \One[i \in A \and j \in B]$ for some $A,B$. Let its “perimiter” $p(M)$ be $|A|+|B|$.
Lemma (Lupanov 1956). Any $n\times m$ boolean matrix can be written as the sum of primitive matrices with total perimeter $O(nm/\log n)$.
By a counting argument (also in Lupanov 1956), this is tight even if the $1$-entries of the primitive matrices are allowed to overlap (sums bigger than $1$ are interpreted as $1$).
Proof
Cut the matrix into $m/k$ “bands” of width $k$. Since there are only $2^k$ possible rows of length $k$, each of these bands can be written as the sum of $\leq 2^k$ primitive matrices, of total width $\leq 2^k \cdot k$ and total height $\leq n$.
Therefore we can write the entire matrix as a sum of primitive matrices with total perimeter
\[(m/k)\cdot(2^k \cdot k + n) = m(2^k + n/k).\]
Setting $k \coloneqq (1-\eps)\log n$ gives total perimeter $nm \cdot \frac{1+o(1)}{\log n}$.