The Riemann curvature tensor for a torsion-free connection is given by:
$$R^d_{cab}V^c=(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d$$
Where $\nabla_a$ and $\nabla_b$ are the covariant derivatives in the $a$ and $b$ directions. But a vector, $V^d$, parallel transported in a direction $a$ has $\nabla_aV^d=0$.
How can the Riemann curvature, which depends on the covariant derivative, measure the effect of parallel transporting a vector around a loop if the covariant derivative of a parallel transported vector is zero?
Edit: changed "vector field" to "vector"
Best Answer
I cannot see the problem. Fix $p\in M$, then $$(R^d_{cab}V^c)_p=((\nabla_a\nabla_b-\nabla_b\nabla_a)V^d)_p$$ implies that $(R^d_{cab}V)_p=0$ if the vector field $V$, defined in a neighborhood $U_p$ of $p$, is such that it is parallely transported along each coordinate curve $x^a$ -- that is $x^k= const$ for $k\neq a$ -- and each coordinste curve $x^b$, defined in $U_p$.
Indeed, in that case e.g. $\nabla_aV^d=0$ holds true in a neighborhood of $p$ so that also the second derivatives in $U_p$, especially $(\nabla_e \nabla_aV^d)_p$, vanish.
It is however by no means obvious that such vector field $V$ exists. That is because it has to satisfy the parallel transport equation along each coordinate curve $x^a$ in $U_p$ with the said fixed a (and b), not only the one exiting $p$!
This is the reason why your argument fails.
To conclude that $(R^d_{cab})_p=0$ (for every value of $a,b,c,d$) along this reasoning, one should prove that there are $n= dim(M)$ vector fields in a neighborhood $U_p$ of $p$ that define a basis of $T_pM$ and such that they are parallely transported along each coordinate curve in $U_p$. Notice that, as a consequence of the parallel transport, these vector fields form a basis at each point of $U_p$.
The point is that the existence of this basis is a quite difficult issue.
Actually, all that is a very known problem and the existence of those $n$ vector fields is not only sufficient for but even equivalent to $Riemann=0$ in a neighborhood of $p$ through the Frobenius theorem.
You can find this statements and its proof in several books (also in Sect. 9.2 Thm 9.18 of these lecture notes of mine which are still work in progress and thus affected by a number of typos).
The issue of the parallel transport along a loop is disentangled with the discussion above where one considers vector fields defined in a full neighbourhood of $p$ and not only on the loop.
In the loop argument one fixes a vector exactly at $p$ and next he/she transports that vector along the loop through $p$. There is no globally defined field in a neighbourhood of $p$ here.
Any attempt to extend the field from the loop to a neighbourhood $U_p$ of $p$ ( which is however impossible as we have two values of $V$ at $p$ in general) does not guarantee that $\nabla_aV^d=0$ in $U_p$ as instead requested above.