Given a coordinate chart, why can we always choose the background metric to be the Euclidean metric determined by the coordinates? Let $(M,g)$ be a Riemannian $n$-manifold. I have one question about the following quote from Geometric Relativity by Dan A. Lee.

It is sometimes convenient to fix a background metric $\bar g$ and compare the geometry to that of $\bar g$. Note that in a single local coordinate chart, one can always choose the background metric to be the Euclidean metric determined by the local coordinates, that is, one can choose $\bar{g}_{ij}=\delta_{ij}$.

In case of possible confusion, let us leave alone the physical meaning of the term background and think of $\bar g$ simply as some other Riemannian metric imposed on $M$ (aside from $g$).
My question is, what is meant by choosing the background metric to be the Euclidean metric determined by the local coordinates? Seemingly, Lee is suggesting that given any coordinate chart $(x_i)$, we can always express the background metric as
$$\bar g=\delta_{ij}dx^i\otimes dx^j.$$
Is it a theorem that can be proved? I used to prove a theorem which states that a Riemannian metric $h$ on $M$ is flat if and only if every point of $M$ is in the domain of a coordinate chart in which the Riemannian metric has the representation
$$h=\delta_{ij}dx^i\otimes dx^j.$$
Are they talking about the same thing? If so, why is the background metric necessarily flat? Isn't $\bar g$ arbitrarily chosen? Isn't there any tiny possibility that we have a Riemannian metric $\bar g$ whose coordinate representation is always
$$\bar g=\delta_{ij}dx^i\otimes dx^j?$$
Thank you.
 A: I believe that they are saying that we can locally construct a positive definite symmetric bilinear form given by
$$
\overline{g} = \delta_{ij} dx^i \otimes dx^j
$$
Hence they are using the term "metric" loosely, as they are not implying that there exists a metric which takes this form in every chart (which would never be true). Rather, for a given chart, there is such a form defined only on this chart that looks just like the restriction of the euclidean metric to this chart.
A: Let $M$ be a manifold, and $\varphi\colon U \to \varphi(U)\subset \Bbb R^n$ be a chart.
It induces coordinates $\{x^1,\ldots,x^n\}$ on $U$, and nothing prevents you from defining the metric $g_U$ on $U$ given by
$$
g_U = \sum_{j=1}^n dx^j\otimes dx^j.
$$
In fact, what you have done is consider the pullback of the euclidean metric through $\varphi$: $g_U = \varphi^* (\mathrm{eucl})$.
If $\varphi\colon U \to \Bbb R^n$ and $\psi\colon V \to \Bbb R^n$ are two chart with an overlap $W:=U\cap V \neq \varnothing$, then you now have two metrics on $W$: $(g_U)|_W$ and $(g_V)|_W$.
There is absolutely no reason for these two metrics to be equal, or even related in any way.
Thus, you cannot say that they define a unique metric on $U\cup V$: they do not even coincide on $W$.
What can be done however is to consider a partition of unity of $U\cup V$ subordinate to the cover $\{U,V\}$, say $\{\lambda_U,\lambda_V\}$, and glue $g_U$ and $g_V$ into $g = \lambda_U g_U + \lambda_Vg_V$.
However, on the overlap $W=U \cap V$, there is no reason that $g$ takes the desired form $\sum_{j} dx^j\otimes dx^j$, in any coordinates system.
You only know that $g$ takes this form on the set $\{x \in U\cup V \mid (\lambda_U(x)=1 \text{ and } \lambda_V(x) = 0) \text{ or } (\lambda_U(x)=0 \text{ and } \lambda_V(x) = 1)\}$.
To go further, there are theoretical obstructions that can be used to show that such a metric almost never exists: such a metric is flat (its curvature identically vanishes), and it is known that many manifolds cannot be endowed with flat metrics.
For instance, among compact orientable surfaces, only the torus admits flat metrics.
