Think of $M_n(F)$ as $V \otimes V^{\ast}$ where $V = F^n$ and $V^{\ast}$ is the dual space. The multiplication map $M_n(F) \times M_n(F) \to M_n(F)$ is a map $(V \otimes V^{\ast}) \otimes (V \otimes V^{\ast}) \to V \otimes V^{\ast}$
which, as it turns out, can be identified with tensor contraction of the two middle terms. It follows that $M_n(F)$, as a left $M_n(F)$-module, can be identified with the tensor product of $V$ (with the natural structure of an $M_n(F)$-module) and $V^{\ast}$ (regarded just as a vector space, the "multiplicity space" of $V$), which is a coordinate-free way of saying that it consists of a direct sum of $n$ copies of $V$ (which in turn is a reformulation of your statement about block diagonals).
Some module theory will put this result in useful context. The above argument shows that $M_n(F)$ is semisimple as a module over itself, hence that it is a semisimple ring, which means any module over $M_n(F)$ is a direct sum of simple modules; moreover, the above argument shows that $F^n$ is the unique simple module of $M_n(F)$ (since any simple module of a ring $R$ appears as a quotient of $R$ regarded as a module over itself), so in fact any module over $M_n(F)$ whatsoever breaks up into a direct sum of copies of $F^n$. This reflects the fact that $M_n(F)$ is Morita equivalent to $F$.