In section 14 of this paper, Einstein affirms:
"For besides the Ricci tensor, there is no tensor of the second rank, which can be built out of the foundamental tensor and their derivatives no higher than the second, and which is also linear in them."
I don't understand the mathematical importance of this tensor:
the combination of terms of the foundamental tensor and their derivatives in the Ricci tensor dosn't seem linear to me, where am I wrong?
could someone explain to me why high order tensors cannot be useful? (is the math solvability the problem?)
By "foundamental tensor," I think Einstein is referring to what we nowadays call a "metric tensor." Roughly speaking, the metric tensor on a manifold is the object that gives a manifold its intrinsic geometry: it determines the lengths of curves, the angles between vectors, whether two geodesics with the same initial direction ever meet, etc.
The idea is to use calculus (i.e., derivatives) to understand this geometry. For example, taking second derivatives of the metric tensor gives rise to the Riemann curvature tensor, which (as the name suggests) measures the curvature of the manifold.
But the Riemann curvature tensor is a rank $4$ tensor, so it is a rather complicated object. In fact, on an $n$-dimensional manifold, the Riemann tensor comprises approximately $\approx \frac{1}{12}n^4$ functions. How can one get a handle on such a long list of functions? One idea is to take a sort of "average." So, by taking the trace (which is like an average) of the Riemann curvature tensor, one obtains the Ricci curvature tensor, which has rank $2$ and consists of about $\approx \frac{1}{2}n^2$ functions. To summarize: the Ricci tensor is essentially the "average curvature" of the manifold.
Since the Ricci tensor is the trace of the Riemann curvature, it is a linear combination of Riemann curvature terms. And since the Riemann curvature is (in a certain sense that I won't make precise) intuitively the "second derivative of the metric tensor," this explains (1).
But what about the third derivative, or the fourth derivative, etc., of the metric tensor? It's a classical theorem that these higher-order derivatives can be expressed in terms of the Riemann curvature tensor and its covariant derivatives, which addresses (2). However, as Deane points out in the comments, third derivatives of the metric tensor can be useful and provide geometric insights.