If $\omega$ is the volumen element of $V$ determined by $T$ and $\mu$, and $f\colon \mathbb{R}^n\to V$ is an isomorphism such that $f^*T=\langle,\rangle$, and such that $[f(e_1),\ldots,f(e_n)]=\mu$, show that $f^*\omega=det$
I am trying to do exercise 4-4 of calculus of manifolds Spivak, The hypothesis of the exercise is very similar to Theorem 4-2.(Calculus of manifolds - Spivak)
what I have tried to give an ortnormal base $[f(e_1),\ldots,f(e_n)]$, such that the pullback $f^*T(e_i,e_j)=\delta_{ij}$, and $f^*\omega(e_1,\ldots,e_n)=\omega(e_1,\ldots,e_n)$.
They are only ideas, since I have not been able to solve it, any help or suggestion is well received
The thorem you attached isn't really what you should be looking at. This exercise amounts to reading carefully the pages on how an inner product and orientation determines a volume element.
Let us briefly recall how this goes. First a definition: if $W$ is an $n$-dimensional real vector space then a volume element is by definition a non-zero alternating $(0,n)$ tensor over $W$; i.e an element $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ (recall that this space is $1$-dimensional so we're essentially choosing a basis for this space).
On an arbitrary vector space $W$ we don't have a specific way of deciding which volume element to take: if we choose one, then any non-zero scalar multiple will do. However, if we have an oriented inner-product space $(W,g,\nu)$ then there is a unique choice, because there is a unique $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ such that for some (and hence any) ordered orthonormal basis $\{w_1,\dots, w_n\}$ of $W$ with the correct orientation $[w_1,\dots, w_n]=\nu$, we have \begin{align} \eta(w_1,\dots, w_n)&=+1. \end{align}
I intentionally avoided the notation $V$ in this discussion because for your question we'll be taking $W=\Bbb{R}^n$, $g=\langle \cdot,\cdot\rangle$ the standard inner product, and the usual orientation $\nu=[e_1,\dots, e_n]$.
For your question, you're given an isomorphism $f$ along with a volume element $\omega$ on $V$. You should know (pretty much by basic definition unwinding) that $f^*\omega$ is an alternating $(0,n)$ tensor on $\Bbb{R}^n$, i.e $f^*\omega\in \mathcal{A}^n(\Bbb{R}^n)$. But of course we already know that $\det$ spans this $1$-dimensional vector space (because as mentioned in the comments this is usually how one defines the determinant... or if you take a more abstract definition of the determinant this also follows easily). Hence, there is a $c\in\Bbb{R}$ such that \begin{align} f^*\omega &= c\cdot \det \end{align} Ideally, we'd like $c=1$, so how do we show that? Simple. Take the standard basis $\{e_1,\dots, e_n\}$ of $\Bbb{R}^n$ then we have \begin{align} c&= c\cdot 1\\ &= c\cdot \det(e_1,\dots, e_n)\\ &=(f^*\omega)(e_1,\dots, e_n)\\ &= \omega(f(e_1),\dots, f(e_n)) \end{align} Now, based on everything I've mentioned above can you justify why this last line equals $1$? (Of course you need to use your assumptions on $f$ at this stage).