An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.
Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles:
$$
T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}.
$$
We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by
$$
\pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n.
$$
For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have
$$
d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}).
$$
where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by
$$
J(x) = \begin{bmatrix}
ie^{ix_1} & 0 & \cdots & 0 \\
0 & ie^{i x_2} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & i e^{i x_n}
\end{bmatrix}.
$$
We can just define
$$
\langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.
We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then
\begin{align*}
\langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\
& = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\
& = \sum_1^n u_j v_j \\
& = \langle u, v \rangle,
\end{align*}
which completes the proof.
Given a linear map $T \colon V \rightarrow W$ between two $n$-dimensional vector spaces, there is no meaningful notion of $\det(T)$ without some additional choices of structure. If you try and define $\det(T)$ to be the determinant of the square matrix $A$ representing $T$ with respect to some choice of bases for $V$ and $W$ then with respect to a difference choice of bases the map $T$ will be represented by the matrix $Q^{-1}AP$ but in general $\det(Q^{-1}AP) = \det(Q^{-1})\det(A)\det(P) \neq \det(A)$ so this is ill-defined.
However, if $V,W$ are real inner product spaces then there is an invariant notion of the absolute value of the determinant of $T$. Namely, define $|\det(T)|$ to be $|\det(A)|$ where $A$ is any square matrix which represents $T$ with respect to some choice of orthonormal bases for $V$ and $W$. If $T$ is represented by $A$ with respect to some choice of orthonormal bases then it will be represented by $Q^{-1}AP$ with respect to a different choice of orthonormal bases but now $P,Q$ are orthogonal matrices and since $\det(P),\det(Q) = \pm 1$ we have
$$ |\det(Q^{-1}AP)| = |\det(Q^{-1})| |\det(A)| |\det(P)| = |\det(A)| $$
so this notion is independent of the orthonormal bases used to represent $T$. Geometrically, on an inner product space one can talk about the volume of $n$-dimensional parallelotopes and if $P$ is such a parallelotope in $V$ then $\operatorname{Vol}(T(P)) = |\det(T)| \operatorname{Vol}(P)$ so $\det(T)$ measures the effect of $T$ on volumes of parallelotopes. This definitions depends on the inner products on $V$ and $W$ and if we change the inner products, $|\det(T)|$ might change. If in addition you endow $V,W$ with an orientation, you can even get rid of the sign ambiguity and define an invariant notion of $\det(T)$ as a real (not neccesarily non-negative) number.
In your case, the Riemannian structure on $M$ gives you an inner product on each tangent space so $Df_{x} \colon T_x M \rightarrow T_{f(x)}M$ is a map between inner product spaces and $|\det(Df_x)|$ is defined as above. Similarly, if you have a subspace $E \subseteq T_x M$ and $Df_{x}$ is an isomorphism then $Df_x|_{E} \colon E \rightarrow Df_x(E)$ is a map between equidimensional inner product spaces (where the inner products are simply the restrictions on $E,Df_x(E)$ of the inner products on $T_xM, T_{f(x)}M$) so $|\det(Df_x|_{E})|$ makes sense. Since you are taking the logarithm of the determinant, I assume that in any case you are interested only in the absolute value of the determinant so this is enough.
Best Answer
If we have an inner product g on a vector space V, we can define an inner product on $\bigotimes_{i=1}^k V$ via $$g(v_1 \otimes \cdots \otimes v_k, w_1 \otimes \cdots \otimes w_k) := \frac{1}{k!}g(v_1, w_1)\cdots g(v_k, w_k).$$
The factor 1/k! has the following explanation: If the wedge product if defined via $$\omega \wedge \eta := \frac{(r+s)!}{r!s!}\operatorname{Alt}(\omega \otimes \eta),$$ where $\omega$ is an r-form and $\eta$ an s-form, then we get $$g(v_1 \wedge \ldots \wedge v_k, w_1 \wedge \ldots \wedge w_k) = \operatorname{det}(g(v_i, w_j)).$$
This gives the nice statement, that if $\{v_1, \ldots, v_n\}$ is an orthonormal basis of V, then $\{v_{i_1} \wedge \ldots \wedge v_{i_k}: 1 \le i_1 < \ldots < i_k \le n\}$ is an orthonormal basis of $\Lambda^k V$.