Annoying sign in geometric interpretation of curvature In this video Susskind gives a heuristic derivation of the curvature formula which I summarize as follows:

In a coordinate system, start with a vector $v_A$ at the origin, and extend it to a parallel vector field along the bottom side of the square (side length $\varepsilon$); let $v_B$ be the value at the bottom right corner. Then extend $v_B$ to a parallel vector field along the right side, obtaining $v_C$, which you similarly extend in parallel fashion along the top side to obtain $v_D$. Finally, you extend $v_D$ again parallel along the left side to obtain $v'_A$.
Now
$$ v_A - v'_A = [(v_C - v_D) - (v_B - v_A)] - [(v_C - v_B) - (v_D - v'_A)] $$
Note $v_C - v_D$ is ($\varepsilon$ times) a covariant derivative of $v$ (to first order) along the $\mu$ direction, and so is $v_B - v_A$. So reasonably we may write (to second order)
$$[(v_C - v_D) - (v_B - v_A)] = \varepsilon^2 \nabla_\nu \nabla_\mu v $$
since we took the difference of $\mu$ covariant derivatives, in the $\nu$ direction. The other term goes mutatis mutandis so that you end up finding (and this is indeed the result Susskind writes down)
$$v_A - v'_A = \varepsilon^2 [\nabla_\nu, \nabla_\mu] v$$
The only problem is I expected to have $[\nabla_\mu, \nabla_\nu]$ in the formula and not the other way around. That is how it's stated in Baez & Muniain's Gauge Fields, Knots and Gravity (p. 247), and also matches the formula obtained in Peskin & Schroeder (p. 484). It's a minor detail but I cannot figure out where this difference in sign could be coming from, and I'd really like to get this right.
 A: Could it be that is just a matter of convention of the definition of the curvature? See https://mathoverflow.net/questions/342288/conventions-for-riemann-curvature-tensor
(I would post this as a comment but don't have enough reputation)
EDIT: Let me give more details. The Riemann curvature tensor can be defined as:
$R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z $.
However, it can also be defined with a global opposite sign in front. It may be the case that Susskind is using the opposite convention as yours.
EDIT#2: now addressing the original question which I misinterpreted at first. I think the problem is not properly tracking second order contributions. First let me say that the RHS of the first equation could be misleading, since you cannot add vectors at different points. Of course we can understand that expression infinitesimally, with all vectors at the original point. Having said this, let me clarify that $v_C-v_D=\epsilon \nabla_\mu v_C$ is only valid up to first order, but since you are looking for a second order result, the correct way is to write
$$v_C-v_D=\epsilon \nabla_\mu v_C -\frac{\epsilon^2}{2} \nabla_\mu\nabla_\mu v_C$$
Then, $v_C$ needs to be written similarly in terms of $v_B$, and $v_B$ in terms of $v_A$  (always up to second order in $\epsilon$). Once this is done,  you get
$$v_A-v'_A=\epsilon^2 [\nabla_\mu,\nabla_\nu]v_A $$
A: I believe I managed to do it following the lines indicated by Alan Garbarz. Let me write it here as it may be of interest.
Instead of following Susskind's argument (which I'm starting to suspect is just wrong), we do one transport at a time, keeping only terms up to quadratic in $\varepsilon$. We start with
$$v_B = v_A + \varepsilon \nabla_\mu v_A + \frac{\varepsilon^2}{2} \nabla^2_\mu v_A$$
Then
\begin{equation*} \begin{split}
 v_C & = v_B + \varepsilon \nabla_\nu v_B + \frac{\varepsilon^2}{2} \nabla^2_\nu v_B \\ & = \left( v_A + \varepsilon \nabla_\mu v_A + \frac{\varepsilon^2}{2} \nabla^2_\mu v_A \right) + \varepsilon \nabla_\nu v_A + \varepsilon^2 \nabla_\nu \nabla_\mu v_A + \frac{\varepsilon^2}{2} \nabla^2_\nu v_A
\end{split} \end{equation*}
This time we'll see some much needed cancellation (the minus sign is since $D$ is in the negative $\mu$ direction from $C$):
\begin{equation*} \begin{split}
v_D & = v_C - \varepsilon \nabla_\mu v_C + \frac{\varepsilon^2}{2} \nabla^2_\mu v_C \\
& = \left( v_A + \varepsilon \nabla_\mu v_A + \varepsilon \nabla_\nu v_A + \frac{\varepsilon^2}{2} \nabla^2_\mu v_A + \frac{\varepsilon^2}{2} \nabla^2_\nu v_A  + \varepsilon^2 \nabla_\nu \nabla_\mu v_A  \right) \\
& - \varepsilon \nabla_\mu v_A - \varepsilon^2 \nabla^2_\mu v_A - \varepsilon^2 \nabla_\mu \nabla_\nu v_A + \frac{\varepsilon^2}{2} \nabla^2_\mu v_A \\
& = v_A + \varepsilon \nabla_\nu v_A + \frac{\varepsilon^2}{2} \nabla^2_\nu v_A + \varepsilon^2 [\nabla_\nu, \nabla_\mu]v_A
\end{split}
\end{equation*}
Finally, we have
\begin{equation*}
\begin{split}
v'_A & = v_D - \varepsilon \nabla_\nu v_D + \frac{\varepsilon^2}{2} \nabla^2_\nu v_D \\
& = \left(v_A + \varepsilon \nabla_\nu v_A + \frac{\varepsilon^2}{2} \nabla^2_\nu v_A + \varepsilon^2 [\nabla_\nu, \nabla_\mu] v_A  \right) \\
& - \varepsilon \nabla_\nu v_A - \varepsilon^2 \nabla^2_\nu v_A + \frac{\varepsilon^2}{2} \nabla^2_\nu v_A \\
& = v_A + \varepsilon^2 [\nabla_\nu, \nabla_\mu]v_A
\end{split}
\end{equation*}
Thus, at last,
$$ v_A - v'_A = - \varepsilon^2 [\nabla_\nu, \nabla_\mu]v_A = \varepsilon^2 [\nabla_\mu, \nabla_\nu]v_A $$
