TITLE: Geometric interpretation of the second covariant derivative
QUESTION [5 upvotes]: I'm having some doubts about the geometric representation of the second covariant derivative.
I know that $\triangledown^{}_a(\triangledown^{}_bv)=(\triangledown^{}_a\triangledown^{}_b)v+\triangledown^{}_{\triangledown^{}_ab}v$
So the Riemann tensor can be defined in two ways :
$R(a,b)v=\triangledown^{}_{a}(\triangledown^{}_{b}v)-\triangledown^{}_{b}(\triangledown^{}_{a}v)-\triangledown^{}_{[a,b]}v\quad$
or
$\quad R(a,b)v=(\triangledown^{}_{a}\triangledown^{}_{b})v-(\triangledown^{}_{b}\triangledown^{}_{a})v$
So far so good (correct me if I'm wrong).
But the problem arises when I read that the Riemann tensor represents the difference between a parallel transported vector along two parts of a closed path. 
Because if $\triangledown^{}_{b}v$ is the parallel transported vector along b, then it stands to reason that $\triangledown^{}_{a}(\triangledown^{}_{b}v)$ is the transported vector along b and then a. But there would be a correction term, so the representation would be inexact.
And conversely, if $(\triangledown^{}_{a}\triangledown^{}_{b})v$ is the transported vector along the path b and then a, the representation would be correct, but I don't understand why...
So, which is correct, A or B ?

Edit : would the difference be linked with the torsion ? Are $\triangledown^{}_{a}(\triangledown^{}_{b}v)$ and $(\triangledown^{}_{a}\triangledown^{}_{b})v$ equal if the connection is torsion free ?
PS : why is the image not integrated in the post ? too large ?

REPLY [2 votes]: After putting all this aside for a while, here's a proposition for a visual answer to my question. Please tell me if something doesn't look right ^^
(note : the vectors are of course random, it's a schematic representation)
Here are the second covariant derivatives and the (contracted) Riemann tensor in the simplest case where $\triangledown^{}_{b}a=\triangledown^{}_{a}b=0$ and $\triangledown^{}_{[a,b]}=\triangledown^{}_{\triangledown^{}_{b}a}-\triangledown^{}_{\triangledown^{}_{a}b}=0$ :

Now in the general case (still torsion-free) with $\triangledown^{}_{[a,b]}v\neq 0$ :

where (it's easier to represent it separately) :

"Not much geometry in all that" ? On the contrary, I think.