Proof of the holonomy principle for $1$-forms Let $M^n$ be a Riemannian manifold with a connection $\nabla$. For a fixed $x\in M$, we define $H:=\text{Hol}_x(\nabla)$ and identify $T_xM\equiv \mathbb{R}^n$. The Holonomy Principle states that:

there is a bijective correspondence between $H$-invariant tensors on $\mathbb{R}^n$ and parallel tensor fields on $M$.

I wasn't able to find any proof I could understand, mainly because of the tensor/bundle dialect, which is still very obscure to me.
I wonder if there is an elementary proof for a particular case, which would shed light on the general idea. For example, if the tensors are $1$-forms, I would have to prove that:

if $\omega_0\in (\mathbb{R}^n)^*$ is $H$-invariant, then there exists $\omega\in\Gamma(TM^*)$ with $\nabla \omega=0$ and $\omega_x=\omega_0$. Conversely, if $\nabla\omega=0$, then $\omega_x$ is $H$-invariant.

Any suggestions?
 A: Constructing $\omega$ from $\omega_0$ is straightforward. We'll need $M$ to be connected though.
For any piecewise smooth path $\gamma : [0,1]\to M$, starting at $x$ and ending at $y$, the parallel transport map $P_\gamma : T_xM\to T_yM$ is an isomorphism of the tangent spaces at $x$ and $y$. Then for any other path $\gamma'$ from $x$ to $y$, if $v\in T_x$, we can produce both the vector $P_\gamma v$ and $P_{\gamma'}v$ in $T_yM$. In order to get a single well defined vector in $T_yM$, these had better be equal.
The condition that $$P_\gamma v = P_{\gamma'}v$$ is equivalent to
$$P_{\gamma'}^{-1}P_{\gamma}v=v,$$
but $$P_{\gamma'}^{-1} = P_{\bar{\gamma}'},$$
where $\bar{\gamma}'$ denotes the reverse path of $\gamma$. Thus $P_{\gamma'}^{-1}P_\gamma v= P_{\bar{\gamma}'\cdot\gamma}v$, where $\cdot$ denotes concatenation of paths. But $P_{\bar{\gamma}'\cdot\gamma}\in H$. Thus we can uniquely parallel transport a vector $v\in T_xM$ to any other point $y$ in $M$ as long as $v$ is invariant under $H$. An identical argument shows that this is true for any other sort of tensor as well.
To be precise. If $v\in T_xM$ is a vector invariant under the action of $H$, then we can extend it to a parallel vector field on $M$, $V$ by defining $V_y = P_\gamma v$, where $\gamma$ is any piecewise smooth path from $x$ to $y$. The same is true for any $H$-invariant tensor at $x$.
For the converse, we'll also stick to vector fields, since everything generalizes identically to tensor fields.
Conversely, suppose we are given a parallel vector field $V$. Let $\gamma$ be a loop at $x$. Since $V$ is parallel, $V|_{\gamma}$ is a parallel vector field on $\gamma$. However $V_x=V_{\gamma(1)}=V_{\gamma(0)}=P_{\gamma}V_{\gamma(0)} = P_{\gamma}V_x$ by definition of the parallel transport map. Thus $V_x$ must be invariant under $P_\gamma$ for any loop $\gamma$ based at $x$. In other words $V_x$ is invariant under the holonomy action.
Edit
To prove this for $1$-forms, the following lemma will be helpful, which basically says that parallel transport of $1$-forms agrees with parallel transport of tangent vectors.

Let $\gamma$ be a piecewise smooth curve in $M$. Parallel transport along $\gamma$ induces an isomorphism $P_\gamma : T_xM\to T_yM$. This produces an isomorphism $P_\gamma^* : T^*_xM\to T^*_yM$ defined by $P_\gamma^* \omega_x = \omega_x \circ P_\gamma^{-1}$. Then $P_\gamma^* \omega_x = \omega_y$ if and only if there is a 1-form field $\omega$ along $\gamma$ such that $\nabla \omega=0$ and  $\omega_0 = \omega_x$ and $\omega_1 = \omega_y$.

Proof.
Suppose we begin with $\omega_x \in T_xM$. Fairly sure the same argument that produces a unique parallel vector field along $\gamma$ extending any tangent vector at the start also applies here producing a 1-form field $\omega$ along $\gamma$ with $\omega_0=\omega_x$ and $\nabla \omega=0$.
It suffices then to prove that $\omega_1 = P_\gamma^*\omega_x$.
Let $u_x\in T_xM$ be arbitrary. Let $u$ be the parallel vector field along $\gamma$. Recall that by definition,
$$ \nabla\omega(u)_t = \frac{d}{dt}\omega(u) - \omega(\nabla_u)_t, $$
but $\nabla u = 0$ and $\nabla\omega = 0$, so $\frac{d}{dt}\omega(u)=0$. Thus $\omega(u)$ is constant. 
In particular, $\omega_y(u_y) = \omega_x(u_x)$, but $u_y = P_\gamma u_x$. Thus
$\omega_y\circ P_\gamma = \omega_x$, since $u_x$ was arbitrary. This rearranges to
$\omega_y = P_\gamma^* \omega_x$, as desired.
The proof for 1-forms
The lemma tells us that it doesn't matter that $H$ is defined by parallel transport of tangent vectors, its action on covectors is compatible with parallel transport of covectors.
Now given an $H$-invariant $\omega_x\in T^*_xM$, we can extend it to $\omega$ on all of $M$ by $\omega_y = P_\gamma^* \omega_x$ for any path $\gamma $ from $x$ to $y$.
Conversely, given $\omega$ with $\nabla \omega = 0$, we have that $P_\gamma^* \omega_x = \omega_x$ for all loops $\gamma$ based at $x$ by the lemma.
