From the comments: "I found that in a seemingly reputable astronomy book. "If it is possible to find a frame in which all the components of g are constant, the space is 'flat'"." --Frank
This question makes sense, but it is phrased in a new-user-hostile way. We have a space $(M,g)$ on our hands. Now consider an open neighborhood $U$ of some point and start picking local frames $(\xi_1, \dots, \xi_n)$ of the tangent space $T_M$ on $U$. The statement in the book says that if you can find a frame such that $g(\xi_j,\xi_k)$ are constant functions on $U$ for all $j,k$, then $g$ is flat on $U$.
The condition says that the $\xi_j$ are parallel with respect to $g$, or that $\nabla \xi_j = 0$, where $\nabla$ is the Levi-Civita connection of $g$. This implies that
$$
R(\xi_j,\xi_k) \xi_l = \nabla_{\xi_j} \nabla_{\xi_k} \xi_l -
\nabla_{\xi_k} \nabla_{\xi_j} \xi_l - \nabla_{[\xi_j,\xi_k]} \xi_l = 0
$$
for all $j, k, l$. Here $R$ is the curvature tensor of $g$. Since $R$ is a tensor on the space of tangent fields and $(\xi_1, \dots, \xi_n)$ is a frame on $U$, this implies that $R = 0$ on $U$, so $g$ is flat.
The difference is all in your head. Literally.
The difference in calling the same object $A$ a "tensor over $\mathfrak{X}(M)$" as opposed to "a tensor field over $M$" is that the former emphasizes the fact that we have an algebraic object: a tensor over some module, while the latter emphasizes the fact that underlying the module there is some manifold and geometry is going on there.
Calling something a tensor field instead of a tensor forces you to remember that $\mathfrak{X}(M)$ is not just some arbitrary module, but that its elements can be identified with smooth sections of the tangent bundle of some manifold. These additional structures are occasionally useful.
Best Answer
A Riemmanian manifold is called flat if its curvature vanishes everywhere. However, this does not mean that this is is an affine space. It merely means (roughly) that locally it "is like an Euclidean space."
Examples of flat manifolds include circles (1-dim), cyclinders (2-dim), the Möbius strip (2-dim) and various other things.
Yet, let me add into the direction of your idea that the universal cover of a complete flat manifold is indeed an Euclidean space.