Short Answer: When considering the metric at a single point, it is a good old fashioned matrix we all know and love from elementary linear algebra.
Long Answer: Well, there is some subtlety to this question, because the entries of a metric tensor in local coordinates $x^{\mu}$ appear to be functions that are sufficiently smooth. Smooth functions form a ring and not a field! So the linear algebra has some intricacies.
This is flawed reasoning! Why? Because we specify the metric on a tangent space $g(-,-)_{p}:T_{p}M\times T_{p}M\to\mathbb{R}$. We just demand that as we "vary $p\in M$ smoothly" that the metric $g$ varies "smoothly" with $p$ (in some sense).
Having made a choice of coordinates $x^{\mu}$ in some neighborhood $U$ of $p$, we have some basis vectors for the tangent space $\left.\mathbf{e}_{\mu}=\partial_{\mu}\right|_{p}$ of $T_{p}M$. These enable us to determine the components of the metric $g_{\mu\nu}=g(\mathbf{e}_{\mu},\mathbf{e}_{\nu})_{p}$ which is a matrix as you noted in (1).
But this is at a single point $p\in M$. The values of $g_{\mu\nu}$ vary smoothly with $p$. Here's the counter-intuitive behaviour of the metric: we write $g_{\mu\nu}(x)$ to indicate the component is a "function" of the coordinates, and it should be smooth "in the obvious way". The problem is that invertibility isn't guaranteed for arbitrary matrices with components being smooth functions! So we work with a very special subcollection of such matrices.