Intuition for curvature in Riemannian geometry

Here's a brutally terse (and slightly imprecise, though "morally correct") account that I hope conveys the geometric content.

In a Riemannian manifold, a tangent vector may be carried uniquely along a piecewise $C^{1}$ path by parallel transport, a geometric notion.

If $X$ and $Y$ are tangent vectors at a point, and if a tangent vector $Z$ is carried around "the small parallelogram with sides $tX$ and $tY$", the parallel-translated vector is, to second order, $Z + t^{2} R(X, Y) Z$. Qualitatively, curvature measures the failure of commutativity of the covariant differentiation operators along $X$ and along $Y$; parallel transporting an orthonormal frame $F$ around a small parallelogram causes $F$ to rotate by an amount linear in each of $X$ and $Y$.

If an (ordered) orthonormal pair $(e_{1}, e_{2})$ is parallel transported around a small square with sides $te_{1}$ and $te_{2}$, it rotates through an angle (approximately) equal to $t^{2}K(e_{1}, e_{2})$. Alternatively, if you're happy with Gaussian curvature as an intrinsic quantity, the sectional curvature of a $2$-plane $E$ at $p$ is the Gaussian curvature of the image of $E$ under the exponential map at $p$.

The Ricci curvature of a unit vector $u$ at $p$ is the average of the sectional curvatures of all $2$-planes at $p$ containing $u$.

The scalar curvature at $p$ is the average of all sectional curvatures of $2$-planes at $p$.

(As in the formulas you give, the averages in the Ricci and scalar curvatures are usually defined and computed as finite averages over suitable pairs of vectors from an orthonormal basis, but in fact the averages may be taken continuously, as described above.)