Inner product and isomorphism From Geometry, topology and physics by Nakahara, page 78:

Let $V=V(m,K)$ be a vector space with a basis $\left\{\mathbf{e}_i\right\}$ and let $g$ be a vector space isomorphism $g:V\to V^*$, where $g$ is an arbitrary element of GL$(m,K)$.

and it then goes on and uses this isomorphism to define the inner product between $a,b\in V$ as $g(a,b)=a^ig_{ji}b^j$.
Why is $g\in\text{GL}(m,K)$? And why is its representation $g:v^j\to g_{ij}v^j$?
 A: There's no need to add confusion by picking a base.
Since $g$ is linear $V\to V^*$, we obtain the bilinear
$$ V\times V\stackrel{(g,\operatorname{id})}\longrightarrow V^*\times V\stackrel{\operatorname{can.}}\longrightarrow K$$
As $g$ is an isomorphim, this bilinear form is non-degenerate: For each $v\ne 0$, $g(v)\ne 0$, hence $g(v)(w)\ne 0$ for some $w$, so $v\cdot w\ne 0$. And for each $v$, there exists $\phi\in V^*$ with $\phi(v)\ne 0$, so $g^{-1}\phi\cdot v\ne 0$. 
A: With my usual notation (from math), my indices are flipped; based on experience, I'm guessing that this is a notational issue with physics, and that objects are operating on the "other" side. So just flip all my indices and that should work. 

Note that $V$ is an $m$-dimensional vector space over $K$. An element of $V^*$ is a map from $V$ to $K$, so it can be seen as a $1\times m$ matrix (relative to a fixed basis). A map from $V$ to $V^*$ relative to this basis will then correspond to $m$ such $1\times m$ matrices, corresponding to the images of the basis vectors. They stack as an $m\times m$ matrix over $K$, and because $g$ is invertible, the rows are linearly independent, so it lies in $\mathrm{GL}(m,K)$.
Thus, every isomorphism $V\to V^*$ corresponds to an $m\times m$ invertible matrix over $K$.  
If $\mathbf{e}_i$ maps to the function $\mathbf{f}_i=(g_{i1},\ldots,g_{in})$, then $\mathbf{f}_i(\mathbf{e}_j) = g_{ij}$. Therefore, if $\mathbf{v}=\sum v_i\mathbf{e}_i$, then $g(\mathbf{v}) =\sum v_i\mathbf{f}_i$, and hence
$$g(\mathbf{v})(\mathbf{e}_j) = \sum_i v_i\mathbf{f}_i(\mathbf{e}_j)= \sum_i v_ig_{ij}$$
which is your equation, but with flipped indices. 
