TL;DR
Question 1 Where did I go wrong in my derivation below?
Question 2 why should including a fifth leg add so much complexity? Is there a more effective way to include the extra leg in the "parallelogram approach"? (which is described below) Since I'm stubborn, I'd still like to know if "my way" works even if there is a cleverer approach.
Question 4 Is actually the case that the Riemann tensor is the 2nd derivative w.r.t "flow time" of EDCBAZ (which is defined below)?
Start of actual Post I've been looking all over the internet for derivations of the Riemann tensor using the "parallelogram approach" that incorporate non-zero torsion, thus making a pentagon since the parallelogram does not close. I haven't found any; the derivations of the Riemann Curvature Tensor \begin{equation} R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z \end{equation} via parallel transport usually fall under two categories:
(1) They usea connection with a vanishing torsion tensor so that $\nabla_{[X,Y]} Z = \mathbf{0}$
(2) They calculate $R(e_i, e_j)e_k$ for a holonomic basis so that $\nabla_{[e_i, e_j]} = \mathbf{0}$
Of course these two approaches are useful and understandable, but I'm stubborn and would like to know how to derive the Riemann tensor (using the following geometrical procedure which I describe) with a connection that HAS torsion (and not defaulting to category 2). I'd like to know where I'm going wrong.
Pick a point $p \in M$ in a Riemannian manifold $(M,g)$. Define a vector $Z$ at $p$ and vector fields $X,Y$ and on this manifold. Parallel transport $Z$ along the flow of $X$ around $p$ for time $t$, ending at point $p_1$ "near" $p$, resulting in a vector $AZ$ ($A$ is the linear map parallel transporting the vector between the associated tangent spaces). Parallel transport $AZ$ along the flow of $Y$ around $p_1$ for time $t$, ending at point $p_2$ "near" $p_1$, resulting in the vector $BAZ$. Parallel transport $BAZ$ along the flow of $-[X,Y]$ for a time $s$, ending at point $p_3$, resulting in $CBAZ$. Parallel transport $CBAZ$ along the flow of $-X$ for $t$ around $p_3$ to $p_4$, giving $DCBAZ$. Finally, parallel transport $DCBAZ$ along the flow of $-Y$ for time $t$ to $p$, giving $EDCBAZ$. I believe it is guaranteed that $EDCBAZ$ coincides with $p$ since we traversed the flow of $-[X,Y]$. Moreover, $s$ is not necessarily equal to $t$ I believe, but it is chosen so that the flow closes the pentagon (which is always possible), and $s \rightarrow 0$ as $t \rightarrow 0$.
Define any smooth vector field $Z^\text{(field)}$ along this pentagon so that $Z^\text{(field)}(p) = Z$, $Z^\text{(field)}(p_1) = AZ$, $Z^\text{(field)}(p_2) = BAZ$, $Z^\text{(field)}(p_3) = CBAZ$, and $Z^\text{(field)}(p_4) = DCBAZ$. I will omit saying "(field)" for brevity, but it should be obvious when we're talking about $Z$ or $Z^\text{(field)}$.
The Riemann tensor is the 2nd derivative w.r.t "flow time" of EDCBAZ: \begin{equation} R(X,Y)Z = \frac{d^2}{dt^2} E(t)(D(t)(C(t)(B(t)(A(t)(Z))))) |_{t=0} \end{equation} So
\begin{equation} R(X,Y)Z = \lim_{h \rightarrow 0} \frac{1}{h^2} \left( E(2h)D(2h)C(2h)B(2h)A(2h)Z - 2E(h)D(h)C(h)B(h)A(h)Z + Z\right) \end{equation}
We expand $A(2h), B(2h), ...$ to 2nd order:
\begin{align*} A(2h) &\approx A(h) + h \nabla_X + h^2 \nabla_X \nabla_X \\ B(2h) &\approx B(h) + h \nabla_Y + h^2 \nabla_Y \nabla_Y \\ C(2h) &\approx C(h) - h \nabla_{[X,Y]} + h^2 \nabla_{[X,Y]} \nabla_{[X,Y]} \\ D(2h) &\approx D(h) - h \nabla_X + h^2 \nabla_X \nabla_X \\ E(2h) &\approx E(h) - h \nabla_Y + h^2 \nabla_Y \nabla_Y \end{align*} The derivatives in these expansions are the covariant derivatives with respect to the (in general non-torsion free) connection since the definition of the covariant derivative is the rate of change of a vector field from its parallel transport. Thus we have to laboriously calculate (for brevity say $E(h) = E, D(h) = D, ...$ henceforth) $$(E - h \nabla_Y + h^2 \nabla_Y \nabla_Y )(D - h \nabla_X + h^2 \nabla_X \nabla_X )...Z - 2EDCBAZ + Z$$ Going through with this, we get \begin{equation} R(X,Y)Z = \lim_{h \rightarrow 0} \frac{1}{h^2} \bigg( Z - EDCBAZ + h \bigg( EDCB \nabla_X Z + EDC \nabla_Y AZ - ED \nabla_{[X,Y]} BAZ - E \nabla_X CBAZ - \nabla_Y DCBAZ \bigg) + h^2 \bigg( EDCB \nabla_{X} \nabla_X Z + E D C \nabla_{Y} \nabla_{X} Z + E D C \nabla_{Y} \nabla_Y A Z - E D \nabla_{[X,Y]} B \nabla_{X} Z - E D \nabla_{[X,Y]} \nabla_{Y} A Z + E D \nabla_{[X,Y]} \nabla_{[X,Y]} B A Z - E \nabla_{X} C B \nabla_{X} Z - E \nabla_{X} C \nabla_{Y} A Z + E \nabla_{X} \nabla_{[X,Y]} B A Z + E \nabla_{X} \nabla_X C B A Z - \nabla_{Y} D C B \nabla_{X} Z - \nabla_{Y} D C \nabla_{Y} A Z + \nabla_{Y} D \nabla_{[X,Y]} B A Z + \nabla_{Y} \nabla_{X} C B A Z + \nabla_{Y} \nabla_Y D C B A Z \bigg) \bigg) \end{equation} Breaking up the limit, using the fact that $A,B,C,D,E \rightarrow \mathbf{I}$ (identity matrix), and simplifying, we have equation (1):
\begin{equation} R(X,Y)Z = \nabla_{X} \nabla_X Z + \nabla_{Y} \nabla_{X} Z + \nabla_{Y} \nabla_Y Z - \nabla_{[X,Y]} \nabla_{X} Z - \nabla_{[X,Y]} \nabla_{Y} Z + \nabla_{[X,Y]} \nabla_{[X,Y]} Z - \nabla_{X} \nabla_{Y} Z + \nabla_{X} \nabla_{[X,Y]} Z + \nabla_{Y} \nabla_{[X,Y]} Z + \lim_{h \rightarrow 0} \frac{Z-EDCBAZ}{h^2} + \lim_{h \rightarrow 0} \frac{1}{h} \bigg( EDCB \nabla_X Z + EDC \nabla_Y AZ - ED \nabla_{[X,Y]} BAZ - E \nabla_X CBAZ - \nabla_Y DCBAZ \bigg) \end{equation} Tackling the first limit here, notice that \begin{equation} Z - EDCBAZ = -\bigg( (EDCBAZ - EDCBZ) + (EDCBZ - EDCZ) + (EDCZ - EDZ) + (EDZ - EZ) + (EZ - Z) \bigg) \end{equation}
So \begin{equation} -\lim_{h \rightarrow 0} \frac{Z-EDCBAZ}{h^2} = \lim_{h \rightarrow 0} \frac{EDCB}{h} \frac{(AZ-Z)}{h} + \lim_{h \rightarrow 0} \frac{EDC}{h} \frac{(BZ-Z)}{h} + \lim_{h \rightarrow 0} \frac{ED}{h} \frac{(CZ-Z)}{h} + \lim_{h \rightarrow 0} \frac{E}{h} \frac{(DZ-Z)}{h} + \lim_{h \rightarrow 0} \frac{1}{h} \frac{(EZ-Z)}{h} \end{equation} Thus \begin{equation} -\lim_{h \rightarrow 0} \frac{Z-EDCBAZ}{h^2} = \lim_{h \rightarrow 0} \frac{1}{h} \bigg(EDCB \nabla_X Z + EDC \nabla_Y Z - ED \nabla_{[X,Y]} Z - E \nabla_X Z - \nabla_Y Z \bigg) \end{equation}
Collecting common terms of $\nabla_X$ and $\nabla_Y$, adding a bunch of $0$'s in a similar manner, and flipping the negative sign, we get equation (2) \begin{equation} \lim_{h \rightarrow 0} \frac{Z-EDCBAZ}{h^2} = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} ( \nabla_X Z + \nabla_Y Z ) + \nabla_X \nabla_X Z + \nabla_Y \nabla_Y Z + \lim_{h \rightarrow 0} \frac{ED \nabla_{[X,Y]} Z}{h} \end{equation}
Inserting this result into equation (1) and simplifying, we get equation (3): \begin{equation} R(X,Y)Z = \nabla_X \nabla_{[X,Y]} Z + \nabla_Y \nabla_{[X,Y]} Z + \nabla_{[X,Y]} \nabla_{[X,Y]} Z + 2 \nabla_X \nabla_X Z + 2 \nabla_Y \nabla_Y Z - 2 \nabla_{[X,Y]} \nabla_X Z - 2 \nabla_{[X,Y]} \nabla_Y Z + \lim_{h \rightarrow 0} \frac{1}{h} \bigg( ED \nabla_{[X,Y]} Z - ED \nabla_{[X,Y]} BAZ + EDCB \nabla_X Z + EDC \nabla_Y AZ - E \nabla_X CBAZ - \nabla_Y DCBAZ \bigg) \end{equation} Remember that $AZ, BAZ, CBAZ, DCBAZ$ are all equal to $Z^\text{(field)}$ at their respective points. Hence, to calculate this last limit, we can cancel the $ED\nabla_{[X,Y]} Z$ terms in the limit and use the same method which got us equation (2). Upon doing this and simplifying once its combined with the rest of equation (3), we get equation (4):
\begin{equation} R(X,Y)Z= -\nabla_X \nabla_Y Z + \nabla_Y \nabla_X Z + \nabla_X \nabla_X + \nabla_Y \nabla_Y Z - \nabla_{[X,Y]} (\nabla_X Z + \nabla_Y Z) + (\nabla_X + \nabla_Y)\nabla_{[X,Y]} Z + \nabla_{[X,Y]} \nabla_{[X,Y]} Z \end{equation}
Here's where I'm stuck. I am unsure whether equation (4) is the same as $\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$. Is there some crazy identity to establish equivalence, or am I just dead wrong? Admittedly, I haven't expanded equation (4) out in component form and cross-examined it with the component form of the correct formula, but why should including a fifth leg add so much complexity? Is there's a more effective way to include the extra leg in the "parallelogram approach"? Since I'm stubborn, I'd still like to know if "my way" works even if there is a cleverer approach. Finally, I am not confident that it is actually the case that the Riemann tensor is the 2nd derivative w.r.t "flow time" of EDCBAZ since I couldn't find any definitive answers online.