"Proof" that zero curvature implies $\partial_a \Gamma^b_{cd}$ is symmetric in $a$ and $c$ I know the claim is wrong. I just want to know where this "proof" goes haywire:
Assume curvature is 0
implies
Parallel transport is path independent
implies
Path integration of Christoffel symbols, $\Gamma ^b_{cd}$, is path independent
implies
$\Gamma^b_{cd}=\partial_c T^b_d$ for some tensor field $T^b_d$
implies
$\partial_a \Gamma^b_{cd}=\partial_a \partial_c T^b_d= \partial_c \partial_a T^b_d=\partial_c \Gamma^b_{ad}$
implies
$\partial_a \Gamma^b_{cd}$ is symmetric in $a$ and $c$
 A: 
Parallel transport is path independent
implies
Path integration of Christoffel symbols, $\tau^b_{cd}$, is path independent
implies
$\tau^b_{cd}=\partial_c T^b_d$ for some tensor field $T^b_d$

The first statement is correct (given the assumptions that precede it). I don't know what the second statement means exactly, but the third statement is definitely not correct. For one thing, the Christoffel symbols are symmetric in their lower two indices (assuming no torsion), but your expression is not (although we could fix that trivially by symmetrizing the indices).$^\star$
What does follow from the first statement is that given two paths, $x_1^\mu(\lambda_1)$ and $x_2^\mu(\lambda_2)$, such that the two paths intersect at points two points, the parallel propagator from one intersection point to the other is the same for $x_1$ and $x_2$.
The parallel propagator $P[x]^\mu_{\ \nu}(\lambda, \lambda_0)$ is a solution of the parallel transport equation satisfied by a vector $V$ parallel transported along a curve $x$
\begin{equation}
\frac{{\rm d} x^\mu}{{\rm d} \lambda} \nabla_\mu V^\nu = 0
\end{equation}
The parallel propagator relates the value of $V^\mu$ at $\lambda$ to its value at $\lambda_0$ after parallel propagation along a curve $x$
\begin{equation}
V^\mu(\lambda) = P[x]^\mu_{\ \ \nu}(\lambda, \lambda_0) V^\nu(\lambda_0)
\end{equation}
We can write a formal expression for the parallel propagator in terms of the path-ordered exponential
\begin{equation}
P[x](\lambda, \lambda_0) = \mathcal{P} \exp\left(\int_{\lambda_0}^\lambda {\rm d \lambda'} A[x](\lambda')\right)
\end{equation}
where
\begin{equation}
A[x]^\mu_{\ \ \nu}(\lambda) = -\tau^\mu_{\nu \sigma} \frac{{\rm d} x^\nu}{{\rm d}\lambda}
\end{equation}
and $\mathcal{P}$ is the path-ordering symbol, which means to expand the exponential in a Taylor series and order the matrix factors $A$ in each term so that each factor of $A$ appears in order of decreasing value of $\lambda$.
Ordinarily, the parallel propagator between two spacetime points will depend on the path taken, but in flat spacetime it will not. Therefore the condition of zero curvature implies that
\begin{equation}
P[x_1]^\mu_{\ \ \nu}(1, 0) = P[x_2]^\mu_{\ \ \nu}(1, 0)
\end{equation}
where $x_1^\mu(0) = x_2^\mu(0)$ is the agreed starting point of the two curves, and $x_1^\mu(1) = x_2^\mu(1)$ is the ending point.
There clearly is some relationship among the Christoffel symbols at different points on the manifold, but it is not as simple as saying that the Christoffel symbols must be the gradient of some function.
The integral equation implied by the equality of the parallel propagators can be converted to a differential equation by considering infinitesimally small loops. In fact this condition will amount to saying that the Riemann curvature is zero
\begin{equation}
R^\mu_{\ \ \nu\rho\sigma} = \partial_\rho\tau^\mu_{\nu\sigma} - \partial_\sigma\tau^\mu_{\nu\rho} + \tau^\mu_{\rho \beta} \tau^\beta_{\nu \sigma} - \tau^\mu_{\sigma \beta} \tau^\beta_{\nu \rho} = 0
\end{equation}
which can be read as a non-linear differential equation for $\tau^\mu_{\nu\sigma}$.
We can in fact check explicitly that your ansatz $\tau^\mu_{\nu\sigma} = \partial_{(\nu} T^\mu_{\ \sigma)}$ does not solve this equation in general (where I've symmetrized your ansatz, and defined the notation $(ab)=\frac{1}{2}(ab+ba)$.
\begin{eqnarray}
R^\mu_{\ \ \nu\rho\sigma} &=& \partial_\rho \partial_{(\nu} T^\mu_{\sigma)} - \partial_\sigma\partial_{(\nu} T^\mu_{\rho)} + \partial_{(\rho} T^\mu_{\beta)} \partial_{(\nu}T^\beta_{\sigma)} - \partial_{(\sigma} T^\mu_{\beta)} \partial_{(\nu}T^\beta_{\rho)} \\
&=& \frac{1}{2} \left( \partial_\rho \partial_\nu T^\mu_\sigma - \partial_\sigma \partial_\nu T^\mu_\rho \right) + O(T^2) \\
&\neq& 0
\end{eqnarray}
where the term in brackets on the last line is not zero in general, and the $O(T^2)$ terms involve two powers of $T$ that cannot cancel the term in brackets.

$^\star$ A good check to do is also whether both sides of an equation transform the same way under coordinate transformations. I'm running out of steam to check if the symmetrized version of your ansatz transforms in the right way to match the transformation of the Christoffel symbols, but if the transformation properties don't match then this is an immediate killer for your identity.

Reference for the parallel propagator stuff: Sean Carroll's GR lecture notes, chapter 3 https://arxiv.org/abs/gr-qc/9712019
