Parallel transport in curved spacetime This is sort of a very introductory question and I am not finding any reference regarding this. And let me know whether my answer is correct or not.
For example, we are parallelly transporting a vector on a surface (intrinsic parallel transport). We take two points and join by a curve and we will do the procedure along the curve. In terms of intrinsic coordinates $P=P(u)$ and $Q=Q(u+du)$. From $P(u)$ if we transport a vector, parallelly, the important condition is
$$
\nabla_i V^j=0\\
\text{or}, \quad \partial_iV^j=-\Gamma^j_{ik}V^k\\
\text{or,} \quad V_\parallel^j(u+du)=V_\parallel^j(u)-\Gamma^j_{ik}V_\parallel^k(u) du^i+\mathcal{O}(du^2)
$$
This transport is done in one path. If we take another path to do the same operation, if $R^a_{bcd} \neq 0$, we will get two different answers for the transported vector component at $Q(u+du)$. So, from the eqn how to interpret this? In particular, there should be some terms which will make $V_\parallel^j(u+du)$ different from one path to another.
Note: These changed $V_\parallel^j(u+du)$ will not make the vector from the point of view of the surface different, since the basis vector also have derivative and have $\Gamma$'s which can cancel the contributions from these $du^n$ terms.
One soln I was thinking, if we take a normal coordinate, for instance, $\Gamma$'s will vanish but $\partial \Gamma$'s will not vanish, hence they actually carry the information of curvature of the manifold. Hence to see actually $V_\parallel^j(u+du)$ are different at different paths, we have to expand upto $(du)^2$ at least.
 A: It's a little tricky to see the path dependence from your equations because the derivative is a local operator. However, if you integrate the condition for parallel transport over a particular path it becomes more obvious. Consider the system below.
Let's work with two coordinates $x^1$ and $x^2$ on a small patch of curved space as seen below. 
If we're given a vector $V$ at point $A$, we can parallel transport it to point $B$ moving in the $x^1$ coordinate direction.
$$\nabla_{e_1}V = 0 \Rightarrow \nabla_1 V^\alpha = 0 \\
\Rightarrow \partial_1 V^\alpha = -\Gamma^\alpha_{1\beta}V^\beta$$
Now integrate both sides along $x^1$ from $A$ to $B$.
$$\int_a^{a+\delta a}\partial_1 V^\alpha \biggr\rvert_{x^2 = b} dx^1 = \int_a^{a+\delta a}-\Gamma^\alpha_{1\beta}V^\beta\biggr\rvert_{x^2 = b}dx^1 \\
\Rightarrow V^\alpha(B) = V^\alpha(A) - \int_a^{a+\delta a}\Gamma^\alpha_{1\beta}V^\beta\biggr\rvert_{x^2 = b}dx^1$$
where $V^\alpha(A)$ are the components of the original vector at point $A$ and $V^\alpha(B)$ are the components of the vector that has been parallel transported to $B$. See we have also taken special care to specify the path as the one on $x^2 = b$. A similar procedure allows you to then parallel transport the vector from $B$ to $C$.
$$V^\alpha(C) = V^\alpha(B) - \int_b^{b+\delta b}\Gamma^\alpha_{2\beta}V^\beta\biggr\rvert_{x^1 = a+\delta a}dx^2 \\
\Rightarrow V^\alpha(C) = V^\alpha(A) - \int_a^{a+\delta a}\Gamma^\alpha_{1\beta}V^\beta\biggr\rvert_{x^2 = b}dx^1 - \int_b^{b+\delta b}\Gamma^\alpha_{2\beta}V^\beta\biggr\rvert_{x^1 = a+\delta a}dx^2$$
I hope by now it's obvious the components of the parallel transported vector are path dependent. For example, had you instead parallel transported $V$ from $A \rightarrow D$ then $D \rightarrow C$, you would obtain a different expression for the components of the parallel transported vector at $C$.
As you hinted in your question, this is closely tied to curvature. In fact, if you parallel transport $V$ around a closed loop, say $A\rightarrow B\rightarrow C\rightarrow D\rightarrow A$, and looked at the difference between the original vector and the parallel transported vector (for a small coordinate patch) the components of the Riemann curvature tensor pop out!
