Let $X$ be an $n$-dimensional complex manifold, let $\omega$ be a Kahler metric on $X$ and let $R$ be the $(4,0)$ curvature tensor of $\omega$. We can simplify the tensor $R$ in different ways, two of which are:

The holomorphic bisectional curvature is $b(\xi,\eta) = R(\xi,\eta,\xi,\eta)/|\xi|^2|\eta|^2$.

The holomorphic sectional curvature is $h(\xi) = R(\xi,\xi,\xi,\xi)/|\xi|^4$.

Now, what exactly is the holomorphic sectional curvature?

To make sense of the question, consider the Ricci curvature of $\omega$. In the Kahler case, this can be defined as the curvature form of the Hermitian metric that $\omega$ defines on the canonical bundle $K_X$. That's a quite nice geometric object that can be interpreted in algebro-geometric ways.

We can make similar sense of the holomorphic bisectional curvature. Consider the projectivized bundle $\pi:\mathbb P(T_X) \to X$. It admits the tautological bundle $\mathcal O(-1) \hookrightarrow \pi^* T_X$ and $\omega$ defines a Hermitian metric on the tautological bundle, whose curvature form at a given point is basically (modulo abuse of notation, sign errors) $$ i\Theta = -|\cdot|^2 \pi^* b + \omega_{\mathrm{FS},\mathbb P(T_X)}, $$ so the negativity of the holomorphic bisectional curvature controls the positivity of $T_X^*$.

By contrast, I know no similar way of thinking about the holomorphic sectional curvature, other than perhaps "the thingy that makes the Schwarz lemma work", which is a much more analytic approach to the situation. It of course has the same average as the Ricci curvature (the scalar curvature), both are controlled by the holomorphic bisectional curvature and neither controls the other, but is that the end of the story? Is the holomorphic sectional curvature a purely analytic object that cannot be attached to any bundle or sheaf?