So when evaluating the length of a curve, say $\gamma$, one uses the arc length paramete $s$: $$\int_{\gamma}ds=\int_{\tau_1}^{\tau_2}\sqrt{\sum_{\mu=1}^{3}(\frac{dx^{\mu}}{d\tau})^2} d\tau$$ which is independent of the parameter $\tau$. We write formally: $$ds^2=(dx^1)^2+(dx^2)^2+(dx^3)^2$$ and $s$ is called the arc length parameter.
Under a coordinate transformation, the line element takes the form: $$ds^2=\sum_{\mu,\nu}g_{\mu,\nu}dx^{\mu}dx^{\nu}$$ where the $g_{\mu,\nu}$ are functions of the coordinates which form a 3$\times$3 matrix.
What I don't understand is how to interpret this under normal Differential Geometry conventions, where the tangent covector $dx$ is simply the dual of the tangent vector $\frac{\partial}{\partial x}$. But what is $dx^{\mu}dx^{\nu}$?
There's three things and depending on context it may mean different things. For example, we can write the metric $g$ (which is a symmetric $(0,2)$ tensor field) as \begin{align} g= g_{\mu\nu}\,dx^{\mu}\otimes dx^{\nu} \tag{$*$} \end{align} Or if we define the "symmetrized tensor product" of two one forms $\alpha,\beta$ as \begin{align} \alpha \beta := \dfrac{\alpha \otimes \beta + \beta \otimes \alpha}{2!}, \end{align} then we can also write (by symmetry of the components $g_{\mu\nu} = g_{\nu\mu}$) \begin{align} g= g_{\mu\nu}\, dx^{\mu}dx^{\nu}. \tag{$**$} \end{align} So far, $(*)$ and $(**)$ express the equality (on some open set) of $(0,2)$ tensor fields (which happen to be symmetric). In other words, for each point $p$ of the manifold $(*)$ and $(**)$ express the equality of bilinear functionals $T_pM \times T_pM \to \Bbb{R}$.
However, when we use the notation $ds^2$, we aren't thinking of a bilinear function, because the notation is very suggestive of the "infinitesimal squared length along a path", and as such, it should be interpreted (pointwise) as a quadratic form $T_pM \to \Bbb{R}$. What is meant is that $ds^2$ is defined by $(ds^2)_p(v):= g_p(v,v)$, for all $p \in M$ and all $v \in T_pM$. So, when we write \begin{align} ds^2 &= g_{\mu\nu}\, dx^{\mu}dx^{\nu} \tag{$** *$} \end{align} the symbol $dx^{\mu}dx^{\nu}$ should be thought of as the quadratic form associated to the bilinear form; i.e \begin{align} (dx^{\mu}dx^{\nu})_p(v) := (dx^{\mu})_p(v) \cdot (dx^{\nu})_p(v); \qquad (p \in M, v\in T_pM) \end{align} where on the RHS we of course have the standard multiplication of real numbers.
Regarding references, for the definition of the symmetrized tensor product, take a look at Lee's Smooth manifolds text (see chapter $12$ on symmetric tensors, and also Chapter $13$ on Riemannian manifolds, pages $314$ and $328$ respectively).
Regarding the interpretation of the $ds^2$ notation as a pointwise quadratic form, see Spivak's Volume $1$ of his Differential Geometry books (in particular page $314$).