# Geometric interpretation of the second covariant derivative

I think you are mixing up (abstract) index notion with an index-free notation. If you take $\xi$, $\eta$ and $v$ to be vector fields, then $(\nabla_\xi\nabla_\eta)v$ does not make sense, since you can apply a covariant derivative only to a vector field and not to the operator $\nabla_\eta$. The equation you are looking for is that if you extend the covariant derivative to an operation on tensor fields, then given $v$, you can form the $\binom11$-tensor field $\nabla v$ defined by $\eta\mapsto \nabla_\eta v$. Applying the covariant derivative to that, one obtains a $\binom12$-tensor field which is usually denoted by $\nabla^2v$. This is indeed given by $(\nabla^2v)(\xi,\eta)=\nabla_\xi\nabla_\eta v-\nabla_{\nabla_\xi\eta}v$. Torsion-freness of the connection then implies that $\nabla_\xi\eta-\nabla_\eta\xi=[\xi,\eta]$ and using this, you see that the curvature is given by $$R(\xi,\eta)(v)=(\nabla^2v)(\xi,\eta)-(\nabla^2v)(\eta,\xi)=\nabla_\xi\nabla_\eta v-\nabla_\eta\nabla_\xi v-\nabla_{[\xi,\eta]}v.$$ Your confusion probably comes from the fact that in (abstract) index notation, one would use $\nabla_av^b$ as a symbol for the $\binom11$-tensor field $\nabla v$. But here $a$ and $b$ are not vector fields but (abstract) indices (and you cannot leave out the "$b$", otherwise $v$ would be a function rather than a vector field). Correspondingly, $\nabla^2v$ will be denoted by $\nabla_a\nabla_bv^c$ (and again $a,b,c$ are indices and not vector fields). In this notation the definition of curvature via $\nabla^2v$ as above then indeed reads as $R_{ab}{}^c{}_dv^d=\nabla_a\nabla_bv^c-\nabla_b\nabla_av^c$.