The issue is the difference between a Riemannian space and a pseudo-Riemannian space. A Riemannian space is one where the metric is positive-definite, ie for all possible vectors X, g(X,X)>0. A pseudo-Riemannian space need not satisfy this, there can be X such that g(X,X)<0 and now the condition is that g is non-singular. The Euclidean metric is Riemannian, it is a diagonal matrix with all +1's down the diagonal, g = diag(+1,+1,+1,...) so g(X,X)>0 for all non-zero X. The Minkowski metric of special relativity is pseudo-Riemannian, since it has (up to arbitrary signature notational choices) the form g = diag(-1,+1,+1,....). Both of these kinds of metric are flat in the Riemann curvature sense, $$R^{a}_{bcd} = 0$$, (the other notion of 'flat' is defined as metrics like the FRW have R = 0 but $$R^{a}_{bcd}\neq 0$$ or the Schwarzchild metric being Ricci flat, $$R_{ab} = R^{c}_{acb} = 0$$, which are weaker conditions than $$R^{a}_{bcd}=0). For metrics in general relativity they have to solve the field equations and can have elaborate forms but thanks to the definition of smooth manifolds meaning that they look flat up close we can use "normal coordinates" to show that any GR metric is also pseudo-Riemannian, as they locally become the Minkowski metric in normal coordinates. A space-like slice of the kinds of pseudo-Riemannian metrics seen in general relativity will give rise to Riemannian metrics on the resultant sub-manifold, as seen in such things as the first and second fundamental forms, which are ways of quantifying curvature on such slices, as well as metric pull backs. But the link doesn't make this restriction, it doesn't mention pseudo-Riemannian anywhere.
Sorry, am I getting a bit too technical for you? None of this is particularly advanced, it's undergrad stuff and since these are just the specifics of things your link brought up there shouldn't be anything wrong with raising the discussion a little, right? After all, you do claim to understand quantum field theory better than Dirac and have done multi-Nobel prize worthy work, a little undergrad stuff which formalised notions of curvature, which you have been 'explaining' to everyone (for years!) shouldn't be an issue, right? Anyway...
As a result of all of this there's numerous mistakes and failures to be precise in that link. It says $$ds^{2} = g_{11}dx^{2} + 2 g_{11}g_{22}dx dy + g_{22} dy^{2}$$ is a Riemannian metric. No, it isn't and for a number of reasons. I'd ask you to give them but I know you'll ignore such a challenge. Firstly we note that the right hand side can be written as $$(g_{11}dx + g_{22} dy)^{2}$$, implying $$ds = g_{11}dx + g_{22}dy$$. Utterly wrong. Utterly. Clearly the person doesn't even know how to use a metric. A metric defines a line element ds by $$ds^{2} = \sum_{a,b}g_{ab}dx^{a}dx^{b}$$. In this case a,b take values 1,2 and we get $$ds^{2} = g_{11}dx^{2} + g_{12}dxdy + g_{21}dydx + g_{22}dy^{2}$$. Since dxdy = dydx and metrics are symmetric so $$g_{12} = g_{21}$$ we get $$ds^{2} = g_{11}dx^{2} + 2g_{12}dxdy + g_{22}dy^{2}$$. Notice the difference? This isn't a perfect square, unlike the link. It involves the off diagonal terms $$g_{12} = g_{21}$$, unlike the link. The link then asserts this is a Riemannian metric, just because he's slapped some arbitrary coefficients in the expression. As I've already explained a Riemannian metric is one which is positive definite, g(X,X)>0 for all X, which amounts to $$g(X,X) = \sum_{a,b}g_{ab}X^{a}X^{b} = g_{11}(X^{1})^{2} + 2g_{12}X^{1}X^{2} + g_{22}(X^{2})^{2} > 0$$. Yes, the expression given in the link is manifestly non-negative since it is, as I just explained, mistakenly a perfect square but that isn't the reason the link asserts the expression is a Riemannian metric....$$