The back of the book provides the following explanation:
If $M = A + iB$, then $\overline{M^t} = \overline{A^t} − i\overline{B^t} = A − iB$.
So we should take $A = \frac{1}{2}(M + \overline{M^t})$ and $B = \frac{1}{2i}(M − \overline{M^t})$, which are of course both Hermitian.
I thought that $\overline{M^t} = \overline{A^t} − i\overline{B^t} = A − iB$ only if $A$ and $B$ are symmetric? So that $A^t$ = $A$ and $B^t = B$.
Then for the second part of the explanation, what makes $A = \frac{1}{2}(M + \overline{M^t})$ Hermitian?
The derivation of the formulas for $A$ and $B$ works after we assume that such hermitian matrices indeed exist. So, it should read like
Now, once we got the formulas, we can prove that they indeed give us hermitian matrices. If
$$A = \frac{1}{2}(M+M^*)$$
then
$$A^* = \frac{1}{2}(M^* + M^{**}) = \frac{1}{2}(M^* + M) = A$$
And similarily for $B$ (note how the signs cancel out):
$$B^* = \frac{1}{-2i}(M^* - M^{**}) = \frac{1}{-2i}(M^* - M) = B$$
NB: I use $A^*$ for $\overline{A^t}$
If you are interested in some more high-level view of the problem, observe that the hermitian transpose operator is self-inverse, and so provides a representation of the group $C_2$ in the space of matrices. General representation theory of $C_2$ tells us that the whole space decomposes into a sum of two subspaces:
The first subspace is the space of hermitian matrices, the second is the space of anti-hermitian matrices.
The only thing that remains is the fact that $A$ is hermitian $\Leftrightarrow$ $iA$ is anti-hermitian.