image of the complement subset of complement of image

If the statement would be true then $H=f(M)\cup f(M)^c=f(M)\cup f(M^c)$.

This states that $f$ is surjective

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Conversely if $f$ is surjective then for every $y\in H$ we can find some $x\in G$ with $y=f(x)$.

If moreover $y\in f(M)^c$ then evidently $x\in M^c$, hence $y=f(x)\in f(M^c)$.

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Final conclusion: the statement is true if and only if $f$ is surjective.