Let $X$ be a smooth $n$-manifold with an oriented atlas $\mathcal U = (U_\alpha, (x_1^\alpha, \dots, x_n^\alpha))$. Let $g$ be a Riemannian metric on $X$.
Let $g_{ij} = g\left ( {\partial \over \partial x_i^\alpha }, {\partial \over \partial x_j^\alpha}\right )$ for some $\alpha$.
If I want to show that $\det g_{ij} >0$ can I argue like this?
(1) By definiton a Riemannian metric is a positive definite quadratic form.
(2) Let $Q$ denote the matrix of the quadratic form. Since the form is positive definite, so is its matrix. Note that the eigenvalues of a positive definite matrix are positive.
(3) Pick a basis in which $Q$ is diagonal. Then the diagonal consists of the eigenvalues and
$$ \det Q = \prod_i \lambda_i > 0$$
since $\lambda_i > 0$.
Yes, more or less. It's worth remembering that the determinant is defined for a linear map $ T: V \rightarrow V $. Strictly speaking, there is no definition for a quadratic form $ V \otimes V \rightarrow \mathbb{R}$.
What you are doing is, choosing a coordinate basis, which induces a non-canonical identification of $ V $ and $ V^* $. With this identification, you can think of $ \tilde{g}: V \rightarrow V $, and then take determinant of $ \tilde{g} $. However the sign of this is a well defined quantity, and independent of the choice of basis.