How scalar curvature of following spacetime can be equal to zero? For an interval of this spacetime,
$$
ds^{2} = c^{2}dt^{2} - c^{2}t^{2}(d \psi^{2} + sh^{2}(\psi )(d \theta^{2} + sin^{2}(\theta )d \varphi^{2})),
$$
scalar curvature is equal to zero. Also, Ricci tensor is equal to zero. How can it be for curved space-time (the metric tensor isn't represented as diag (1, -1, -1, -1))? 
Addition.

 A: You are incorrect to suppose that this spacetime is curved. In fact, up to some conditions on the coordinate ranges, this is simply a piece of Minkowski spacetime. Let me put it in this form:
$$ds^2 = dt^2 - t^2(d\psi^2 + \sinh^2\psi\,d\Omega^2)\text{,}$$
where $d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2$ is the metric for a unit $2$-sphere, and we can go from spherical Minkowski form $dT^2 - dR^2 - R^2\,d\Omega^2$ to yours by the substitution
$$\begin{eqnarray*}R = t\sinh\psi &\quad\quad& T = t\cosh\psi\end{eqnarray*}$$
Simply, the Ricci tensor is zero because $R_{\mu\nu} = 0$ is a coordinate-invariant condition: if a tensor zero in some coordinate chart (e.g., Minkowski), then it is zero in every coordinate chart (e.g., yours). Naturally, the Riemann tensor is also zero.
Interestingly, this is the also the hyperbolic FLRW ("Big Bang") metric in the limit of zero density, where a galaxies don't interact gravitationally but every galaxy sees itself a center of cosmological expansion. For details, see the Milne universe model.


...the article at the link below says that this interval corresponds to an open flat (Space curvature = -1/R^2 < 0) expanding Universe (look at the picture in the question). Can you explain that?

Sure: spacetime curvature is not the same thing as space curvature, or more generally the curvature of a manifold is not the same thing as the curvature of a submanifold of lower dimensionality. Take Euclidean space $E^3$ in spherical coordinates, $dS^2 = dR^2 + R^2d\Omega^2$, and look at submanifolds of constant radius:
$$d\sigma^2 = R^2(d\theta^2 + \sin^2\theta\,d\phi^2)\text{,}\quad R = \text{constant}$$
One can verify by direct computation that the scalar curvature is positive, as one might expect from them being spheres. Obviously, if you 'slice up' $E^3$ by planes, each will have zero curvature instead.
In your case, what's going on is just the same thing, except we're 'slicing up'/'foliating' the Minkowski spacetime $E^{1,3}$ by spacelike hypersurfaces of the form
$$d\sigma^2 = t^2\left[d\psi^2 + \sinh^2\psi\,d\Omega^2\right]\text{,}\quad t = \text{constant}$$
In this form, it's not hard to guess that the curvature for each hypersurface is same as above except negative. (Let me know if you need help calculating it explicitly, but for the moment I think intuitively motivating the fact that this makes sense would be more fruitful.)
Since coordinate changes do not change the scalar curvature, we can let $\rho = t\sinh\psi$ to get $d\rho = t\cosh\psi\,d\psi$ and
$$t^2\,d\psi^2 = \frac{d\rho^2}{\cosh^2\psi} = \frac{d\rho^2}{1 + \sinh^2\psi} = \frac{d\rho^2}{1+\rho^2/t^2}\text{,}$$
putting it in the same form as (80) except for variable names; the other part of the metric should be more obvious.
A: Firs of all, the metric depends on the specific coordinate system. If metric is not diagonal, it does not mean that the space is curved -- just take some crazy coordinates in the flat space and you will have a really messy metric. Futhermore, only vanishing Riemann tensor ensures that there are local coordinates in which metric is diag(1,-1,-1,-1), that is, the space is locally flat.
(Pseudo-)Riemann manifolds with vanishing Ricci tensor are called Ricci-flat, and they are the solutions of Einstein equations in empty space-time without cosmological constant. For example, Schwarzschild solution is actually Ricci-flat (except the singularity at the origin).
A: The point is that saying "spacetime is curved" does not necessarily mean anything specific about the metric or any of the curvature tensors. I think for physics (meaning, for general relativity), "spacetime is curved" should mean that the scalar curvature is not zero, since that's what shows up in the action. Of course, what matters is the variation of the scalar curvature with the metric (which gives the Einstein equations), so even if the scalar curvature is locally zero, it does not need to be zero everywhere. The metric can look complicated if you choose some complicated coordinates, so just looking at the metric is not enough to determine if "spacetime is curved."
For my money, flat spacetimes have $R=0$ everywhere, and Ricci-flat spacetimes have zero Ricci tensor everywhere (and no matter). These are the most common and most precise ways of considering a spacetime to be flat (or curved).
