Is "X is a flat Riemannian manifold" equivalent to "for any metric g on X, there is a change of coordinates that can transform g to a tensor with only constants on its diagonal"?
[Math] Flat manifold and metric
differential-geometrymanifolds
Related Solutions
A good place to start is the wikipedia article which describes how various curvatures transform under conformal change a Riemannian metric. Now, apply the formula for the transformation of the full curvature tensor to your situation, taking into account that initially you have a flat metric.
I think this question deserves an answer here in physics SE, since these notions are very common in Relativity.
For a manifold $M$ equipped with a metric (either Euclidean or Lorentzian) $g$, locally flat means that every point $p\in M$ admits an opens set $U \ni p$ endowed with coordinates $$\psi : U \ni q \mapsto (x^1(q),\ldots, x^n(q)) \in \psi(U) \subset \mathbb R^n$$ such that the metric in these coordinates takes the constant canonical form $$g_{ab}(q) = \delta_{ab}\quad \mbox{Riemannian case}\:,$$ or $$g_{ab}(q) = \eta_{ab}\quad \mbox{Lorentzian case}\:.$$ The point is that there is no guarantee that there is such a $U$ which covers the whole manifold. For this reason we use the adverb locally in front of flat. An example is a two dimensional cylinder in $\mathbb R^3$ equipped with the metric induced by the standard metric on $\mathbb R^3$.
For a manifold $M$ equipped with a metric (either Euclidean or Lorentzian) $g$, globally flat means that there is a global coordinate system $$\psi : M \ni q \mapsto (x^1(q),\ldots, x^n(q)) \in \psi(M) \subset \mathbb R^n$$ such that the metric in these coordinates takes the constant canonical form $$g_{ab}(q) = \delta_{ab}\quad \mbox{Euclidean case}\:,$$ or $$g_{ab}(q) = \eta_{ab}\quad \mbox{Lorentzian case}\:.$$ In this case $(M,g)$ is isometrically identified with an open portion (possibly the whole space) of $\mathbb R^n$ equipped with the standard (Euclidean or Lorentzian) metric.
In view of the definitions above, the answer to your question (i) depends on the extension of your coordinate system. If it covers the whole manifold we have a globally flat manifold. If it is not the case, we can only say that a region of the manifold, the one covered by the coordinates, is flat.
Flat space and Euclidean space have more or less the same meaning. They usually indicate Minkowski spacetime (i.e. a spacetime isometrically isomorphic to $\mathbb R^n$ equipped with the standard Minkowsky metric $\eta_{ab}$) or $\mathbb R^n$ equipped with the standard Euclidean metric $\delta_{ab}$. However are a bit vague terms whose meaning (if global or local) should be understood from the context.
Regarding your last question. If all $C_i$ are positive, the metric can be recast into the standard form $\delta_{ij}$ simply rescaling (changing) the coordinates with a factor $1/\sqrt{C_i}$. If some constant is negative, with a similar procedure the metric can be reduced into a constant form $h_i \delta_{ij}$ where $h_i = \pm 1$, which is Lorentzian only if all $h_i$ are $+1$ but one (or all $h_i$ are $-1$ but one depending on the adopted convention). It is impossible that a $C_i$ vanishes because it would imply $det [g_{ij}]=0$ which is not admitted by definition of (non-degenerate) metric.
ADDENDUM. An apparently related question concerns the so called locally flat coordinates around a given point. This is an independent issue. These coordinates are also called Riemannian normal coordinates or Gaussian normal coordinates. Are coordinates $x^1,\ldots, x^n$ defined in a neighborhood of a point $p\in M$ such that exactly at $p$ the components of the metric have their canonical form (e.g. $g_{ab}(p)=\delta_{ab}$ or $g_{ab}(p)=\eta_{ab}$ depending on the nature of the metric) and $\frac{\partial g_{ab}}{\partial x^c}|_p =0$. It is possible to prove that every manifold equipped with a smooth metric admits such a coordinate system for every given point $p\in M$.
It is also possible to extend the definition referring to a geodesic $\gamma\subset M$ instead of a point $p$: Exactly on $\gamma$ the components of the metric have their canonical form (e.g. $g_{ab}(\gamma)=\delta_{ab}$ or $g_{ab}(\gamma)=\eta_{ab}$ depending on the nature of the metric) and $\frac{\partial g_{ab}}{\partial x^c}|_\gamma =0$. This form gives rise to a very precise mathematical description of Einstein's equivalence principle in GR when $\gamma$ is a timelike geodesic.
Best Answer
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.