show/hide this revision's text 2 clarification about what S is doing

Here is one way to think about your first question which at least might provide a more geometric picture about what is going on.

I want to think about the curvature $R(X,Y)$ as parallel transport around the infinitesimal parallelogram $X \wedge Y$. If I drag a vector $Z$ around the parallelogram $X \wedge Y$, the result is $R(X,Y)Z$. Since the connection is metric, the map $Z \mapsto R(X,Y)Z$ is actually an infinitesimal rotation; this is the observation that $$\langle R(X,Y)Z, W\rangle = -\langle Z, R(X,Y)W\rangle$$

Now I want to define a new operator $S$ which acts bilinearly on pairs of 2-vectors. This will be $$S(X\wedge Y, Z \wedge W) = \langle R(X,Y)Z,W\rangle$$ where I have summed over some basis of 2-planes in $\bigwedge^2 T_pM$. Geometrically, $S$ is reports how much the operation of dragging an infinitesimal 2-plane $Z \wedge W$ rotates as it is dragged around the 2-plane $X\wedge Y$. To see that this is well-defined we need only to check $S(-,Z\wedge W) = -S(-, W \wedge Z)$. But this follows precisely because of the previous equation for $R$.

From here on, I'm going to use the metric to think of $S$ as $$S(X \wedge Y) = \sum_{2\text{-planes } Z\wedge W} \langle R(X,Y) Z,W \rangle ~Z\wedge W$$ The somewhat mysterious "pair swap" symmetry $\langle R(X,Y) Z, W\rangle = \langle R(Z,W)X,Y\rangle$ can now be interpreted as saying that the operator $S$ is symmetric. In particular, this means that we can take the spectral decomposition of $S$ to get a basis of orthogonal unit-area eigenplanes $X_i \wedge Y_i$, $$S(X_i \wedge Y_i) = \lambda_i \cdot X_i \wedge Y_i$$ The eigenvalues $\lambda_i$ are your sectional curvatures for this basis; any other sectional curvatures can be easily computed from these.

Note that knowledge of $S$ is now clearly sufficient to reconstruct the curvature tensor, since $$\langle R(X,Y)Z, W\rangle = \langle S(X\wedge Y), Z \wedge W \rangle$$ so in fact the sectional curvature tensor $S$ determines the usual curvature tensor $R$.
show/hide this revision's text 1

Here is one way to think about your first question which at least might provide a more geometric picture about what is going on.

I want to think about the curvature $R(X,Y)$ as parallel transport around the infinitesimal parallelogram $X \wedge Y$. If I drag a vector $Z$ around the parallelogram $X \wedge Y$, the result is $R(X,Y)Z$. Since the connection is metric, the map $Z \mapsto R(X,Y)Z$ is actually an infinitesimal rotation; this is the observation that $$\langle R(X,Y)Z, W\rangle = -\langle Z, R(X,Y)W\rangle$$

Now I want to define a new operator $S$ which acts bilinearly on pairs of 2-vectors. This will be $$S(X\wedge Y, Z \wedge W) = \langle R(X,Y)Z,W\rangle$$ where I have summed over some basis of 2-planes in $\bigwedge^2 T_pM$. Geometrically, $S$ is the operation of dragging an infinitesimal 2-plane $Z \wedge W$ around the 2-plane $X\wedge Y$. To see that this is well-defined we need only to check $S(-,Z\wedge W) = -S(-, W \wedge Z)$. But this follows precisely because of the previous equation for $R$.

From here on, I'm going to use the metric to think of $S$ as $$S(X \wedge Y) = \sum_{2\text{-planes } Z\wedge W} \langle R(X,Y) Z,W \rangle ~Z\wedge W$$ The somewhat mysterious "pair swap" symmetry $\langle R(X,Y) Z, W\rangle = \langle R(Z,W)X,Y\rangle$ can now be interpreted as saying that the operator $S$ is symmetric. In particular, this means that we can take the spectral decomposition of $S$ to get a basis of orthogonal unit-area eigenplanes $X_i \wedge Y_i$, $$S(X_i \wedge Y_i) = \lambda_i \cdot X_i \wedge Y_i$$ The eigenvalues $\lambda_i$ are your sectional curvatures for this basis; any other sectional curvatures can be easily computed from these.

Note that knowledge of $S$ is now clearly sufficient to reconstruct the curvature tensor, since $$\langle R(X,Y)Z, W\rangle = \langle S(X\wedge Y), Z \wedge W \rangle$$ so in fact the sectional curvature tensor $S$ determines the usual curvature tensor $R$.