[Added 7.10.2018:]
Due to public demand, let me specify explicitely that this is an answer to the following question:
If we can show that parallel
transport depends only on the homotopy class of the path, then we can
produce a covariantly constant section by parallel transporting some
fixed vector to each point. But, I am not sure how to get this as a
consequence of 0 curvature. Any suggestions?
[end added content]
Here is a proof: We pick a homotopy $h_s(t) = H(s,t)$ between two paths $h_0$ and $h_1$ in $M$ and consider the parallel transport along each of the curves $h_s$. We can then show that the resulting vector field along $H$ will be parallel, not only in the $t$-direction, but also in the $s$ direction. This step uses the vanishing curvature assumption. From this the claim follows, because a homotopy leaves the endpoints fixed. Therefore the vector field – being parallel in the $s$-direction at $t=1$ – must be constant there, if we only vary $s$. Here are the details.
Let $I= [0,1]$. Suppose we are given a homotopy $$H: I\times I \to M, \quad (t,s)\mapsto H(t,s) = h_s(t)$$ between two paths $h_0 = H(\cdot,0)$ and $h_1 = H(\cdot,1)$. Let $x = h_s(0)$ and $y=h_s(1)$ denote the two endpoints (which are assumed to be fixed during the homotopy). Let $v\in E$ with $p(v) = x$. We denote by $t\mapsto V(s,t)$ the parallel transport of $v$ along $t\mapsto h_s(t)$ for any $s\in I$, i.e. $V$ satisfies
$$\nabla_{t} V = 0 \;\; \forall s,t\in I \quad \text{and} \quad V(s,0) = v\;\; \forall s\in I.$$
Vanishing curvature implies that $\nabla_t \nabla_s V = \nabla_s \nabla_t V$ for all $s,t\in I$. Thus, for any fixed $s\in I$, we have that $t\mapsto \nabla_s V(s,t)$ satisfies $\nabla_t \nabla_s V(s,t) = \nabla_s\nabla_t V(s,t) = 0$ and $\nabla_sV(s,0) = \nabla_s v = 0$. By uniqueness of parallel transport, it follows that $\nabla_s V = 0$. In particular, for $t=1$, we obtain that $\nabla_s V(s,1) = 0$. Since $h_s(1) = y$ is the constant path at $y$, it follows that $V(0,1) = V(1,1)$. So the parallel transport of $v$ along $h_0$ and $h_1$ agrees at their common endpoint. $\square$
Let $(M,g)$ a Riemannian manifold with $\nabla$ any $g$-compatible connection, then $\frac{g(V(t),V(t))}{dt}=g(\nabla_{\dot c} V(t),V(t))+g(V(t),\nabla_{\dot c}V(t))=0$ so $g(V(t),V(t))$ is constant so $P$ preserves lengths.
Edit: see comments.
You have $0=\nabla_Vg$ which is the same as $0=\sum_{i,j} \nabla_V(g_i\otimes g_j)=\sum_{i,j}(\nabla_Vg_i\otimes g_j+g_i\otimes \nabla_Vg_j)$ where $g_i,j$ are 1-forms. Then we get: $$(\nabla_Vg)(X,Y)=\sum_{i.j}\nabla_Vg_i(X)g_j(Y)+g_i(X)\nabla_Vg_j(Y))$$ now by the rule of covariant derives in 1-forms we get $(\nabla_Vg_i)(X)=V(g_i(X))-g_i(\nabla_VX)$ by putting this in the previous formula we get:
$$\sum_{i,j} V(g_i(X))g_j(Y)-g_i(\nabla_V(X))g_j(Y)+g_i(X)V(g_j(Y))-g_i(X)g_j(\nabla_V(Y))=V(g(X,Y)-(g(\nabla_VX,Y)+g(X,\nabla_VY))$$
Best Answer
@John Ma's answer is a good one, but I'd also like to point out that parallel vector fields are even rarer that Killing vector fields. For example, on a round sphere, there are plenty of Killing vector fields but no nontrivial parallel fields. Basically, the existence of a parallel vector field is equivalent to the condition that the metric splits locally into a Riemannian product of a one-dimensional manifold and an $(n-1)$-dimensional one. This implies, in particular, that the sectional curvatures of planes containing $V$ are all zero.