In their original form, the Einstein field equations simply said the following,

$$ T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T = \frac{1}{\kappa}R_{\mu\nu} $$

where $g_{\mu\nu}$ is the covariant metric tensor, which is the metric expressed in the tangent basis, and $T$ is the quantity $g_{\mu\nu}T^{\mu\nu}$, where Einstein summation is used. $\kappa$ is a constant.

This was the original form of the Einstein field equations, which express a relation between the information proferred by the stress energy tensor, or rather a complicated polynomial thereof on the left and the Ricci tensor, which has a very concrete interpretation as the link will tell, you on the right. By some rather trivial algebra, one can recover from these equations, the form of the Einstein field equations you have mentioned in the question.

There probably is no meaning to the components of the Einstein tensor, which is simply obtained by subjecting the original form of the field equations to some arguably trivial algebra.

In their original form, the Einstein field equations simply said the following,

$$ T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T = \frac{1}{\kappa}R_{\mu\nu} $$

where $g_{\mu\nu}$ is the covariant metric tensor, and $T$ is the quantity $g_{\mu\nu}T^{\mu\nu}$, where Einstein summation is used. $\kappa$ is a constant.

This was the original form of the Einstein field equations, which express a relation between the information proferred by the stress energy tensor, or rather a complicated polynomial thereof on the left and the Ricci tensor, you on the right. By some algebra, one can recover from these equations, the form of the Einstein field equations you have mentioned in the question.

There probably is no meaning to the components of the Einstein tensor, which is simply obtained by subjecting the original form of the field equations to some algebra.

As the others have mentioned in the comments above, there really is no clear way to answer or arguably even interpret this question appropriately, but the following is a somewhat rigorous explanation of what is really going on.

A full course of differential geometry clearly cannot be offered here, but the barebones explanation is the following. In general relativity, spacetime is basically interpreted as a four-dimensional manifold with a metric of indefinite signature $(1,3)$. What this basically means is that spacetime is a collection $M$ of points and a collection $\mathcal{U}$ of subsets of $M$ such that given a subset $U \in \mathcal{U}$, there exists a subset $E$ of $\mathbb{R}^{n}$ and an infinitely differentiable map $\phi : U \rightarrow E$ such that the map $\phi$ is both invertible and the inverse is infinitely differentiable (a.k.a smooth) as well.

This manifold is also equipped with a metric structure. This means that if you have two points $p,q\in M$, then there is a function $d: M\times M \rightarrow \mathbb{R}$ such that,

1. $d(p,q)\leq d(p,r) + d(r,q)$
2. $d(p,q) = 0$ iff $p=q$

Normally, positivity is demanded as well, but that condition is relaxed in Einstein's theory, to account for causality, which I won't talk about here, but you can visit https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold to understand that in depth. Now, at each point, one may define the so called tangent space $T_p M$, which again, I cannot describe in detail, so ref. https://en.wikipedia.org/wiki/Tangent_space for the same. Assuming you know some linear algebra, one has a natural isomorphism between $T_pM $ and $T^{**}_p M$. Elements of $T^*_pM$ are called contravariant tensors and elements of $T^{**}_pM$ are called covariant vectors, defined at the point $p$. Tensors now may be naturally defined analogously in terms of the Cartesian products of the tangent space with itself. As a concrete example, a covariant 2-tensor at a point $p$ is an element of $(T_pM\times T_pM)^{**}$.

Now, coming to your question, in covaruant form, the stree-energy tensor is simply a rank two covariant tensor that roughly speaking, encodes properties about the local distribution of matter and energy. The truth of the matter is that in their original form, the Einstein field equations simply said the following,

$$ T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T = \frac{1}{\kappa}R_{\mu\nu} $$

where $g_{\mu\nu}$ is the covariant metric tensor, which is the metric expressed in the tangent basis, and $T$ is the quantity $g_{\mu\nu}T^{\mu\nu}$, where Einstein summation is used. $\kappa$ is a constant. $R_{\mu\nu}$ here is the Ricci tensor, for which in favour of brevity, I simply offer the following reference https://en.wikipedia.org/wiki/Ricci_curvature.

This was the original form of the Einstein field equations, which express a relation between the information proferred by the stress energy tensor, or rather a complicated polynomial thereof on the left and the Ricci tensor, which has a very concrete interpretation as the link will tell, you on the right. By some rather trivial algebra, one can recover from these equations, the form of the Einstein field equations you have mentioned in the question.

To cut, rather meaninglessly at this point, a long story short, there probably is no meaning to the components of the Einstein tensor, which is simply obtained by subjecting the original form of the field equations to some arguably trivial algebra.

In their original form, the Einstein field equations simply said the following,

$$ T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T = \frac{1}{\kappa}R_{\mu\nu} $$

where $g_{\mu\nu}$ is the covariant metric tensor, which is the metric expressed in the tangent basis, and $T$ is the quantity $g_{\mu\nu}T^{\mu\nu}$, where Einstein summation is used. $\kappa$ is a constant.

This was the original form of the Einstein field equations, which express a relation between the information proferred by the stress energy tensor, or rather a complicated polynomial thereof on the left and the Ricci tensor, which has a very concrete interpretation as the link will tell, you on the right. By some rather trivial algebra, one can recover from these equations, the form of the Einstein field equations you have mentioned in the question.

There probably is no meaning to the components of the Einstein tensor, which is simply obtained by subjecting the original form of the field equations to some arguably trivial algebra.