May be $e(x)$ a frame vector at the point $x \in M$ for a manifold $M$. Given the following connection equation:
$d_Xe+e_{|X_+}-e_{|X_-}= d_Xe + J_Xe = 0$.
Here, $d_X$ denotes the directional derivative in the direction of the vector field $X$ and $J_X$ is the jump operator; it measures how the frame vector (more mathematically the section of the tangent bundle) jumps when parallel transporting along $X$. I have to compute the corresponding curvature tensor. Here is what I have done:
Deriving this expression by $Y$ yields:
$d_Yd_Xe+(d_YJ_X)e+J_X(d_Ye)=d_Yd_Xe+(d_YJ_X)e-J_XJ_Ye=0$. (A)
If $X$ and $Y$ are interchanged, it holds:
$d_Xd_Ye+(d_XJ_Y)e-J_YJ_Xe=0$. (B)
Assuming that the vector fields $X,Y$ are chosen such that the Lie bracket is $[d_X,d_Y]=0$ it follows by subtraction of (A) from (B):
$d_Xd_Ye+(d_XJ_Y)e-J_YJ_Xe-(d_Yd_Xe+(d_YJ_X)e-J_XJ_Ye)=(d_XJ_Y-d_YJ_X+J_XJ_Y-J_YJ_X)e := R_{XY}e = 0$.
The curvature tensor $R_{XY}$ looks similar to the ordinary affine (or fiber bundle) curvature tensor, but reformulated with evaluating the jump operator it has the form:
$(d_XJ_Y-d_YJ_X+J_XJ_Y-J_YJ_X)e=(J_{Y|X_+}-J_{Y|X_-}-J_{X|Y_+}+J_{X|Y_-})e + J_X(e_{|Y_+}-e_{|Y_-})-J_Y(e_{|X_+}-e_{|X_-})=(J_{Y|X_+}e-J_Ye_{|X_+})-(J_{Y|X_-}e-J_Ye_{|X_-})-(J_{X|Y_+}e-J_Xe_{|Y_+})+(J_{X|Y_+}e-J_Xe_{|Y_+})$.
Defining $W_{X|Y}:=(J_{X|Y}e-J_Xe_{|Y})$ with $e=e_{|Y}-\delta_Y e,J_{X|Y}=J_X+\delta_Y J_X$ it follows $W_{X|Y}= - J_X (\delta_Y e) + (\delta_Y J_X)e - (\delta_Y J_X) (\delta_Y e)$ (the higher order (h.o.) is $-(\delta_Y J_X) (\delta_Y e)$). It also follows $\delta_{Y_+} e - \delta_{Y_-} e = J_Ye$; hence:
$R_{XY}e=W_{Y|X_+}-W_{Y|X_-}-W_{X|Y_+}+W_{X|Y_-}=-J_YJ_Xe+J_XJ_Ye+J_XJ_Ye-J_YJ_Xe+h.o.=2[J_X,J_Y]e+h.o.$.
Moreover $J_XJ_Ye=-J_Xd_Ye$ and if the vector fields $X,Y$ are constant (e.g. basis vectors of $R^n$) it also follows $J_XJ_Ye=d_Xd_Y$ and finally $[J_X,J_Y]e=0$. The remaining contribution to the curvature in this case is the higher-order term. However, this can only be assumed for the vector $e$!
Question: Are my calculations right?