The 'right' way to express the line element $ds^2$ Taking Minkowski space as an example, the line element is commonly expressed as,
$$ds^2 = dt^2 - dx^2-dy^2-dz^2.$$
However, when expressing the line element in terms of an orthonormal basis $e^a$ it may be expressed as,
$$ds^2 = e^t \otimes e^t - e^x \otimes e^x - e^y \otimes e^y - e^z \otimes e^z.$$
In the case of Minkowski space, we of course have $e^t = dt$ and so forth. This then makes me question whether in the first line element $dt$ is somehow different from the $e^t = dt$ since there is no tensor product symbol. As I understand it, $e^t$ is a differential form and so lives in a section of the cotangent bundle of the underlying manifold.
What's the correct way to express the line element?
 A: *

*An (orthonormal) basis $e_a$ is not a collection of differential forms, but a collection of vector fields. One would denote its dual basis (of the differential 1-forms) by differential forms $\theta^a$ if it is a non-coordinate basis and by something like $\mathrm{d}x^a$ if it is a coordinate basis. The dual basis is defined by $\theta^a(e_b) = \delta^a_b$.

*The metric is canonically defined as a 2-tensor acting on vector fields, i.e. for a basis of differential 1-forms $\theta^a$ it has components
$$ g = g_{ij}(\theta^i\otimes\theta^j).$$
In case of a coordinate basis, 
$$ g = g_{ij}(\mathrm{d}x^i\otimes\mathrm{d}x^j)$$
and by convention one often writes $(\mathrm{d}x^i)^2$ for $\mathrm{d}x^i\otimes\mathrm{d}x^i$. For the Minkowski metric and the standard coordinate basis this clearly reproduces your first formula.

*In general, there is no orthonormal coordinate basis (this would mean you're in flat space). However, non-coordinate bases can often be found that are orthonormal. Suppose you have such an $e_a$ with a dual basis $\theta^a$. Since an orthonormal basis has $g(e_a,e_b) = \eta_{ab}$, where $\eta = \mathrm{diag}(1,-1,-1,-1)$ is the usual Minkowski metric, you have that $g_{ab} = \eta_{ab}$ in the dual basis, i.e.
$$ g = \theta^1\otimes\theta^1 - \sum_{i=2}^4 \theta^i\otimes\theta^i,$$
which reproduces your second formula.
Note that all of these manipulations require careful consideration what geometric object you are dealing with: The vector fields $e_a$ are different from the differential forms $\theta^a$, and writing $g = e_a\otimes e_a$ would be non-sensical - the metric tensor by definition is a tensor acting on vectors, not a tensor made as the tensor product of vector acting on differential forms.
A: I'm not entirely sure the question but with regards to defining a line element $(ds)^2$ in terms of basis vectors we have the relation $g_{ij}=\mathbf{e}_i \cdot \mathbf{e}_j$ so the line element is given by 
$$ (ds)^2  = g_{ij} dx^i dx^j = (\mathbf{e}_i \cdot \mathbf{e}_j )dx^i dx^j$$
