Your first and biggest misconception is that you seem to be mixing up real and complex dimensions. A Riemann surface is a $2$-dimensional real manifold, but as a complex manifold it is $1$-dimensional. Hence you have just one local complex coordinate on $M$, namely $z$. The local coordinates are not $(z, \bar{z})$.
There is more structure on the tangent bundle of $M$ and the bundle of $k$-forms on $M$. Here's the general picture. Let $X$ be a $2n$-dimensional complex manifold, and consider local coordinates $(x_1, \dots, x_n, y_1, \dots, y_n)$. Using the complex structure on the tangent bundle (which we denote $i$) we can locally define vector fields
$$\frac{\partial}{\partial z_j} = \frac{1}{2} \left(\frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right),$$
$$\frac{\partial}{\partial \bar{z}_j} = \frac{1}{2} \left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right).$$
We also locally have the dual basis of $1$-forms
$$dz_j = dx_j + i ~dy_j,$$
$$d\bar{z}_j = dx_j - i ~dy_j.$$
Then we get a splitting of the cotangent bundle
$$T^\ast M = \Lambda^{1,0} T^\ast M \oplus \Lambda^{0,1} T^\ast M,$$
where pointwise $\Lambda^{1,0} T^\ast M$ is spanned by the $dz_j$ and $\Lambda^{0,1} T^\ast M$ is spanned by the $d\bar{z}_j$. More generally, we have the splitting
$$\Lambda^k T^\ast M = \Lambda^{k,0} T^\ast M \oplus \Lambda^{k-1, 1} T^\ast M \oplus \cdots \oplus \Lambda^{1, k-1} T^\ast M \oplus \Lambda^{0,k} T^\ast M,$$
where for $p + q = k$, $\Lambda^{p,q} T^\ast M$ is spanned pointwise by the $dz_{j_1} \wedge \cdots \wedge dz_{j_p} \wedge d\bar{z}_{j_{p+1}} \wedge \cdots \wedge d\bar{z}_{j_k}$. Sections of $\Lambda^{p,q} T^\ast M$ are called $(p,q)$-forms, sections of $\Lambda^{k,0} T^\ast M$ are called holomorphic $k$-forms, and sections of $\Lambda^{0,k} T^\ast M$ are called antiholomorphic $k$-forms.
On a Riemann surface, the canonical bundle is the bundle of holomorphic $1$-forms, i.e.
$$K = \Lambda^{1,0} T^\ast M.$$
Therefore in terms of a local coordinate $z$, a section $\alpha$ of $K$ is described by
$$\alpha(z) = f(z) ~dz$$
for some holomorphic function $f$. Your expression of the form
$$\omega = F ~dz \wedge d\bar{z}$$
would locally describe a section of $\Lambda^{1,1} T^\ast M$, not a holomorphic $1$-form.
Your local description is closer to being ok. Here I would let $\{e_1, \dots, e_n\}$ be a local frame for $E$, i.e. a set of $n$ local sections that form a basis for each fiber $E_z$ of $E$ for $z$ in our coordinate neighborhood. Then $\{\phi_1, \dots, \phi_n\}$ would be the dual frame defined by
$$\phi_i(z)(e_j(z)) = \delta_{ij}$$
for all $z$ in our coordinate neighborhood. Then locally a section $\sigma$ of $E^\ast$ would look like
$$\sigma(z) = \sum_{k = 1}^n g_k(z) ~\phi_k(z) \tag{$\ast$}$$
for each $z$ in our coordinate neighborhood.
Putting the above together, a section of $K \otimes E^\ast$ is a linear combination of sections of the form $\alpha \otimes \sigma$, where $\alpha$ is a section of $K$ and $\sigma$ is a section of $E^\ast$, and locally such a $\alpha \otimes \sigma$ looks like
$$(\alpha \otimes \sigma)(z) = \sum_{k = 1}^n f(z)g_k(z) ~dz \otimes \phi_k(z)$$
for each $z$ in our coordinate neighborhood.
If $TM$ is the full holomorphic tangent bundle of $M$, then a section $\alpha \otimes \sigma$ of $K \otimes E^\ast$ can be considered as a map from sections of $TM \otimes E$ to the space of holomorphic functions on $M$ as follows. Let $\alpha \otimes \sigma$ have the form $(\ast)$ determined above. A section of $TM \otimes E$ is a linear combination of sections of the form $v \otimes s$, where $v$ is a section of $TM$ and $s$ is a section of $E$. Locally we have
$$(v \otimes s)(z) = \left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \otimes \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right).$$
Then we have
\begin{align*}
(\alpha \otimes \sigma)(v \otimes s)(z) & = \alpha(v)(z) \sigma(s)(z) \\
& = f(z) ~dz\left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \sum_{k = 1}^n g_k(z) ~\phi_k(z) \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right) \\
& = f(z) \left( a(z) \cdot 1 + b(z) \cdot 0 \right) \sum_{k = 1}^n \sum_{j = 1}^n \delta_{kj} g_k(z) h_j(z) \\
& = f(z)a(z) \sum_{k = 1}^n g_k(z)h_k(z).
\end{align*}
When the above is considered over a single point, we see how to map an element of $TM \otimes E$ to $\Bbb C$ using an element of $K \otimes E^\ast$.
The first thing you mention is that, if you fix a basis for $V$, you get a non-canonical isomorphism $V\cong V^*$. Similarly, fixing a basis for $V^*$, you get a non-canonical isomorphism $V^*\cong (V^*)^*=V^{**}$. The "magic" is that, when you compose these two isomorphisms, you get an isomorphism $V\cong V^{**}$ which of course depends on your choice of bases, but which is exactly equal to the canonical isomorphism
$$\iota: V\to V^{**};\quad \iota(v)=\iota_v,$$
regardless of what choices you made to define $\Phi$ and $\Phi^*$.
The two extra equations you give show that
$$\Phi^*\circ\Phi = \iota.$$
What may confuse you is that, in order to show these two maps are equal, we choose bases. But this is normal: we need these bases to define $\Phi$ and $\Phi^*$ to start with. But note that the two extra equations hold regardless of what bases were chosen.
Best Answer
Yes, if you have a vector bundle $E \to M$ with a fiber metric $h$, then there is a natural isomorphism induced by the usual identification of a vector space with its dual in the presence of an inner product (not necessarily of definite signature): The map $\Phi: E \mapsto E^*$ is given fiberwise by $$\Phi_p: v \mapsto h_p(v, \,\cdot\,),$$ which is manifestly independent of basis, and as a section of $\text{Hom}(E, E^*)$ it obviously is as smooth as $h$ is, because it is $h$, viewed dually.
(Of course, all of this requires a metric or an appropriate substitute, in general there is no natural isomorphism between a vector space and its dual, or between a vector bundle and its dual.)