Some questions about the hodge star operator. I am a beginner of Hodge Theory. When I read the notes, I find the following which makes me feel uncomfortable:

Below is an example:

First, from my understanding, $V$ is a vector space and $V^{*}$ is its dual space. Usually, we should use $dx, dy, dz$ to represent an element in dual. From my understanding, the example use $dx, dy, dz$ to represent elements in $V$. What's wrong here? Should I make confusion? 
Second, I also want to ask if it is necessary to define an inner product in $V^{*}$. Usually, in a manifold, we can define a (0,2) tensor $g$ such that $g_{ij}=g(\partial_{i},\partial_{j})$. However, how to define an inner product for its dual? i.e. $g(dx^{i},dx^{j})$; is it simply the inverse of $(g_{ij})$? How to derive it?
Third, I do not understand the volume form. Why there are exactly two choices differing by sign?
I hope someone can answer the above questions to clear my mind about Hodge theory.
 A: The first excerpt you give talks about the Hodge star for an abstract vector space $V$ which has a metric, i.e. a bilinear function $V\times V\to \mathbb R$. For the second excerpt, you set $V = T_p^*M$ for a point $p$ in a Riemannian manifold $M$. Thus the elements of $V$ are $dx,dy,dz,$ etc. Then, as your second question asks, you need a metric on the cotangent space, $\tilde g: T_p^*M\times T_p^*M\to \mathbb R$, which is induced by the metric $g:T_pM\times T_pM\to \mathbb R$. 
This can be defined a few ways. One way is to use the metric dual of vectors in $T_pM$: If $X = X^i\partial_i$, then define a 1-form dual to $X$ by $\alpha_X = g(X,\cdot)= g_{ij}X^i dx^j$. Then $\tilde g$ on $T_p^*M$ is defined so that $X\to \alpha_X$ is an isometry, i.e.
$$ g(X,Y) = \tilde g(\alpha_X,\alpha_Y).$$
So in particular, if $\tilde g^{ij}:= \tilde g (dx^i, dx^j)$,
\begin{align*}
g_{ij}X^iY^j &= \tilde g^{kl}(g_{ki}X^i)(g_{lj}Y^j) \quad\text{for all $X,Y$}\\
\implies g_{ij} &= \tilde g^{kl}g_{ki}g_{lj}
\end{align*}
So by using the inverse metric with $g_{ki}g^{im} = \delta_k^m$,
\begin{align*}
g_{ij}g^{im}g^{jn} &= \tilde g^{kl}g_{ki}g_{lj}g^{im}g^{jn}\\
g^{mn} &= \tilde g^{mn}
\end{align*}
As you mentioned, it is the inverse of $g$. You could define it this way from the beginning too.
For your last question, the volume form is a choice of vector Vol in $\Lambda^n V$ which has length 1 (i.e. $|\text{Vol}|^2 := g(\text{Vol},\text{Vol}) = 1$). Since $\Lambda^n V$ is 1 dimensional, the space is just a line, and so you have only two choices.
I hope that answers everything.
A: Basis for the dual space It's not true that we typically write the elements of the dual of a vector space $dx, dy, \ldots$. In any case, either there is interstitial material missing or the example leaves several things unsaid. I'll attempt to fill in the gaps:
In our setup, we are working on the manifold $\Bbb R^2$, which comes equipped with the standard coordinates $(x, y)$ and the standard metric $g$. Then, $dx$ and $dy$ are the coordinate $1$-forms (sections of the cotangent bundle $T^* \Bbb R^2$) given by taking the exterior derivative of $x$ and $y$, respectively.
At any point $(x, y) \in \Bbb R^2$, we take inner product space $V$ in the description of the Hodge star operator to be the cotangent space $T_{(x, y)}^* \Bbb R^2$ equipped with the dual metric $g^{-1}$. The resulting Hodge star operation $\ast_{(x, y)}$ is thus a map $T_{(x, y)}^* \Bbb R^2 \to T_{(x, y)}^* \Bbb R^2$.
Now, since we have a Hodge star map at each point, we can assemble those maps into a bundle map, which we just denote $$\ast : T^* \Bbb R^2 \to T^* \Bbb R^2 .$$ In particular, it acts at each point by $$\ast (dx\vert_{(x, y)}) = dy\vert_{(x, y)} , \qquad \ast (dy\vert_{(x, y)}) = - dx \vert_{(x, y)} .$$ Viewing $\ast$ as a tensorial map on sections, $\Gamma(T^* \Bbb R^2) \to \Gamma(T^* \Bbb R^2)$, we get the indicated formulas $$\ast dx = dy, \qquad \ast dy = -dx .$$
The dual metric For any inner product $g$ on a (real or complex) vector space $V$, there is a unique corresponding inner product $g^{-1}$ on $V^*$, called the dual metric. Each can be characterized in terms of the other by $$\operatorname{tr}(g^{-1} \otimes g) = \operatorname{id}_V ,$$ where the left-hand side of the equation just indicates that we contract of an index of $g^{-1}$ with an index of $g$.
Unwinding definitions lets us make this more concrete: If we choose a basis $(e_a)$ of $V$, we can represent the inner product $g$ as the matrix $[g]$ with $(a, b)$ entry $[g]_{ab} = g_{ab} = g(e_a, e_b)$. Then, the matrix representation of $g^{-1}$ with respect to the basis $(\epsilon^a)$ of $V^*$ dual to $(E_a)$ is just the inverse of $[g]$, that is, $$[g^{-1}] = [g]^{-1} .$$ The $(a, b)$ entry of is $[g^{-1}]^{ab} = (g^{-1})^{ab} = g^{-1}(\epsilon^a, \epsilon^b)$.
The volume form The definition of volume form in the text is not standard. (Usually, a volume form on an $n$-dimensional vector space $V$ is defined to be a nonzero element of $\bigwedge^n V^*$). In any case, $g$ determines an inner product on $\bigwedge^n V$, which by slight abuse of notation we'll also denote $g$. But $\bigwedge^n V$ is $1$-dimensional, so if we pick any nonzero element $\Omega \in \bigwedge^n V$, we can write any element of $\bigwedge^n V$ as $\lambda \Omega$. By bilinearity,
we have $$g(\lambda \Omega, \lambda \Omega) = \lambda^2 g(\Omega, \Omega) ,$$ and so there are precisely $2$ volume forms $\operatorname{Vol}$ that satisfy $$g(\operatorname{Vol}, \operatorname{Vol}) = 1.$$ Explicitly in terms of our choice of $\Omega$, $\operatorname{Vol} = \pm \lambda \Omega$, where $\lambda := g(\Omega, \Omega)^{-1 / 2}$.
