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One thing the scalar curvature forgets is all the information about the coordinates you were using. The notion of having a "big Ricci curvature" is one that can only be defined in a particular coordinate system. For instance, in coordinates $(x,y)$, $R_{xx}$ and $R_{yy}$ could be big, but in some other set of coordinates $(u,v)$, $R_{uu}$ and $R_{vv}$ could be small or even zero. Because the scalar curvature is a scalar, its value is coordinate-independent.
I assume you had Riemannian spaces in mind, but there are some very well-motivated examples in relativity. For example, when Schwarzschild originally wrote down the metric for the vacuum region surrounding a spherically symmetric body, the metric had a singularity at a certain radius $r>0$, which would be external to the body if the body was very compact. In the coordinates he was using, the singularity was present in the Riemann tensor. However, decades later it was discovered that the singularity could be removed by switching to different coordinates. A hint of this nonphysical character of the singularity was that there was no singularity in any scalar measure of curvature. The scalar curvature $R^{ab}R_{ab}$ was automatically zero because the Einstein field equations require the Ricci tensor to be zero in a vacuum. However, there are other scalar measures of curvature such as Kretschmann invariant $R^{abcd}R_{abcd}$, and these also vanished.
The Kretschmann invariant does blow up at $r=0$ in the Schwarzschild solution, and this is now interpreted as the physical singularity at the center of a black hole.