Why does this falsely lead to $g_{\mu \nu}$ always being the identity map? So I posted a question about the tetrad basis but later realised that there is a more fundamental, underlying question that is better suited here.
I’m using abstract index notation to denote all tensors, meaning that all Latin indices are symbolic references to slots and all Greek indices are references to specific components in a specified basis. Tensor products are written as two tensors next to each other, for example: $T^{ab} = v^a w^b$, and contractions are written by repeating indices, for instance $R_{ab} = R_{acb}^c$.
Essentially, given a metric tensor $g = g_{ab}$ and its inverse tensor $g^{-1} = g^{ab}$, the compound tensor $g_a^b = g_{ac} g^{cb}$ that you get from inserting one metric into either slots of the second one is said to be the identity map:
$$
g_a^b = g_{ac} g^{cb} = \delta_a^b.\tag{1}
$$
In practice, this means that $g^b_a$ applied to any tensor just changes the name of the slot from $a$ to $b$ or the other way around (i.e. it “does nothing”). The components are of course given by:
$$
g_\mu^\nu = g_{\mu \sigma} g^{\sigma \nu}.\tag{2}
$$
So the components $g^\nu_\mu$ in any basis may differ from the components $g_{\mu \nu}$.
But we also have, from inserting the basis vectors into the tensor to get its tensor components:
$$
g_{\mu \nu} = g_{ab} \left( \frac{\partial}{\partial x^{\mu}}\right)^a \left( \frac{\partial}{\partial x^{\nu}}\right) ^b = (dx^\mu )_a \left( \frac{\partial}{\partial x^{\nu}}  \right)^a = \delta_\nu^\mu\tag{3}
$$
where we have used that $g_{ab} T^a = T_b$ for any tensor, and hence also for basis vectors $T^a = \left(\frac{\partial}{\partial x^\mu}\right)^a$. Moreover, I assumed that lowering an index for a basis vector yields its corresponding dual vector, where “corresponding” is tested via the condition $$(dx^\mu )_a \left( \frac{\partial}{\partial x^{\nu}}  \right)^a \equiv dx^\mu \left( \frac{\partial}{\partial x^{\nu}}  \right) = \delta^\mu_\nu .\tag{4}$$
As you can see, here is the problem. First of all, why do I end up with mixed indices on either sides of the equation? Secondly, why do I get that the components of the metric in any basis is the identity map?
 A: 
I assumed that lowering an index for a basis vector yields its corresponding dual vector

This is incorrect. The metric does add a "normalization factor" to the expression. For example, consider the metric for the 2-sphere,
$$\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta \mathrm{d}\phi^2.$$
Notice that
$$g_{ab}\left(\frac{\partial}{\partial \phi}\right)^b = \sin^2\theta \left(\mathrm{d}\phi\right)_a,$$
which is necessary in order to have
$$g_{\phi\phi} = g_{ab}\left(\frac{\partial}{\partial \phi}\right)^a\left(\frac{\partial}{\partial \phi}\right)^b = \sin^2\theta.$$
Of course, the case of a non-diagonal metric would also bring in mixed terms, making the expression more complicated, but I think this example suffices to show the problem. While this is enough to show that Eq. (3) is wrong, the index matching was bothering me as well. The example of Minkowski metric shows that the index should be lowered, though. Indeed,
$$\eta_{ab}\left(\frac{\partial}{\partial t}\right)^b = - \left(\mathrm{d}t\right)_a = - \left(\mathrm{d}x^0\right)_a = \left(\mathrm{d}x_0\right)_a.$$
With these things in mind, notice that the correct computation would be
\begin{align}
g_{\mu\nu} &= g_{ab}\left(\frac{\partial}{\partial x^\mu}\right)^a\left(\frac{\partial}{\partial x^\nu}\right)^b, \\
&= g_{\mu\rho}\left(\mathrm{d} x^\rho\right)_b\left(\frac{\partial}{\partial x^\nu}\right)^b, \\
&= g_{\mu\rho}\delta^{\rho}{}_{\nu}, \\
&= g_{\mu\nu},
\end{align}
which is a trivial result, but what would really be surprising is if we found the components of the metric tensor without choosing the metric tensor.
