For Riemannian manifolds, what does it mean to express a vector using the dual basis? I've read most of Lee's Smooth Manifolds and also his Intro to Riemannian Manifolds. In Lee's books, for a manifold $M$, he defines a tangent vector $v \in T_p M$ as a derivation at a point $p$, i.e., a linear map that maps real-valued functions $f \in C^\infty (M)$ to a real number: $v: C^\infty (M) \rightarrow \mathbb R$. And I'm not sure if it's necessarily has to be of this form, but he seems to identify them earlier by the fact that a tangent vector $v$ yields a map $D_v|_p: C^\infty (M) \rightarrow \mathbb R$, the directional  derivative of a function $f$.
He also defines tangent covectors as linear functionals on the tangent space at a point $p$, i.e., if the tangent space is a vector space $V$, then the covector space $V^*$ is the set of linear maps $\omega: V \rightarrow \mathbb R$, and the basis $E_j$ and dual basis $\epsilon_i$ of the vector/covector spaces are orthogonal in the usual way: $\langle  \epsilon_i , E_j \rangle = \epsilon_i (E_j) = \delta^i_j$.
That all makes sense to me, but I'm conceptually confused by other things I saw. E.g., in An Elementary Introduction to Information Geometry (p5) by Frank Nielsen, the author says:


and:

I read this to mean that a given vector $v$ can be expressed using either its primal basis ($e_i$) or its dual basis ($e^i$), in Einstein notation:
$$v = v^i e_i = v_i e^i$$
What I don't understand is how $v$ can be expressed with either basis, given that the basis of the primal space are vectors, while the basis of the dual space are covectors! the way I read it in the Lee books were that vectors and covectors are different entities, which do different things.
My best guess is that it somehow involves using the isomorphism the metric provides between a basis and its dual, e.g., $e^i = g^{ij} e_j$, but I don't see how that gets around the fact that $v$ still must be a vector in the end.
How can a vector be expressed using both the primal or dual bases?
edit: and with regards to the last picture, I'm even more confused: my understanding (and what he says) is that the metric $g$ is a mapping $V \times V \rightarrow \mathbb R$, and that defines an isomorphism that basically allows us to translate between vectors and covectors, i.e., $g: V \times V \rightarrow \mathbb R \Rightarrow g: V \rightarrow (V \rightarrow \mathbb R) \Rightarrow g: V \rightarrow V^\ast$. And that's what he appears to be doing in eq's 7 and 8, but it's translating between $e_i$ and $e^{\ast j}$, which seem to be both vectors...
 A: First of all, Lee is a pure mathematician, while Nielsen is an applied mathematician. As it frequently happens with applied mathematicians, Nielsen is loose with terminology. He also made one typo in the quoted passage. His use of notation is also different from Lee's (whose notation are more standard).
In more detail: Nielsen does not explain which space the objects called $e^{*i}$ belong to, but, from the context, they belong to the vector space $V$ (the tangent space) and not $V^*$ (the cotangent space). I assume, he also uses (elsewhere in the book?) genuine covectors $e^i\in V^*$. In his picture, however, he forgot to put $*$ in $e^i$ (this would be a typo). The relation between $e^{*i}\in V$ and $e^i\in V^*$  then would be that
$$
\langle e^{*i}, v\rangle= e^i(v)$$
for every $v\in V$. However, it is also possible that he simply refuses to use actual covectors (differential 1-forms) in the book and leaves it unexplained what the symbols $dx^i$ stand for (if he ever uses differential forms in the book).
He notes (correctly) that when
$$
v=\sum_i v^i e_i= \sum_i v_ie^{*i},
$$
one can recover the coefficients $v^i$ and $v_i$ using the inner product in $V$:
$$
v^i= \langle v, e^{*i}\rangle
$$
$$
v_i= \langle v, e_i\rangle
$$
Edit. Regarding your question in the edit. The Riemannian metric defines "musical" isomorphisms between tangent and cotangent spaces
$$
\flat: T_pM\to T^*_pM, \sharp: T^*_pM\to T_pM. 
$$
(One should not call these isomorphisms $g$.) What he (Nielsen) is using is that $\sharp(e^i)=e^{*i}$ and not that $\sharp(e^{*i})=e_{i}$ (the latter would be nonsense indeed). Here, I am using the notation $\{e^i: i=1,...,n\}$ for the basis of $V^*$ which is dual to the basis $\{e_i: i=1,...,n\}$ of $V$. Admittedly, he never introduced the notation $e^i$, but this is what he is implicitly using.
