I am reading page 28 of the 1994 version of Principles of Algebraic Geometry by Griffith. Let $M$ be a complex manifold of dimension n, Griffith defined a Hermitian metric to be a positive definite Hermitian inner product $($ , $)_z:T'_z(M)\otimes \overline{T'_z(M)}\to \mathbb{C}$ depending smoothly on $z$, meaning that for local coordinates $z$ on $M$ functions of the form $h_{ij}(z)=(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})_z$ are smooth. Denoting the hermitian metric by $ds^2$, then $ds^2=\sum\limits_{i,j} h_{ij}(z)dz_i\otimes d\overline{z_j}$. I get how this decomposition works, and I inferred that $(T'_z(M)\otimes \overline{T'_z(M)})^*$ must be an $n^2$ dimensional complex vector space. So, I don't get why there exists an $n$-tuple of forms $(\phi_1,...,\phi_n)$ of type $(1,0)$ so that $ds^2=\sum\limits_i \phi_i\otimes\overline{\phi_i}$. Griffith also mentions that $($ , $)_z$ induces an inner product on ${T^{*}}_{z}'(M)$, the dual vector space of $T'_z(M)$, what would be that inner product?
PS: here is the original text
