Let $M$ be a manifold, $E$ a vector bundle on $M$. Write ${\rm Fr} \, (E)$ for the frame bundle. The following are equivalent:
- a horizontal distrbution on ${\rm Fr} \, E$
- equivariant parallel transport on ${\rm Fr} \, E$
- Linear parallel transport on $E$
So all we need to do is check that linear paralell transport on $E$ is the same as specifying $\nabla_X s$. Suppose that we have linear paralell transport. Then we can define
$$ (\nabla_X s)_m = \lim_{t \to 0} \frac{T(s_{\gamma(t)})-s_m}{t} $$
where $ \gamma $ is the integral curve for $X$ at $m$ and $T(s_{\gamma(t)})$ is the paralell transport of $s_{\gamma(t)}$ to $m$ along $ \gamma^{-1} $. This defines a covariant derivative. On the otherhand, if we have a covariant derivative and a curve $ \gamma : I \to M$, then we have a covariant derivative on $ \gamma^* E $ defined by
$$ \nabla_T s := \nabla_{\gamma'T} s $$
(recall that $ \gamma' : TI \to \gamma^* TM $ is a map of vector bundles.) This formula makes sense because sections of a pullback bundle are germs of sections over the image. But vector bundles on $I$ are smoothly trivial, so $ \gamma^* E = I \times V $ for some vector space $V$. It follows from the fundamental theorem of ODEs that there is a paralell section $ s(t) \in V $ defined for $t \in [0,\epsilon) $. Using Gronwall's inequality, this section can't blow up in finite time, so it must be defined on all $[0,1]$. This gives us paralell transport. We still need to check that these constructions are inverse to each other. SInce the covariant derivative and paralell transport are determined locally, we just need to check when $M = \mathbb{R}^n$. If you start with a system of parallel sections and define a covariant derivative $\nabla$, then your original sections are paralell for $\nabla$. For the other direction we need to do some work.
Let $V = \mathbb{R} \{ e_1,\dots,e_d \} \cong \mathbb{R}^d $ be a vector space and consider the vector bundle $\mathbb{R}^n \times V$ over $ \mathbb{R}^n$. Suppose that $\nabla$ is a connection on $V$. Then we have that
$$ \nabla_{\partial / \partial x_k} e_j = \sum_i \Gamma_k^{ij} e_i $$
Consider the curve $ \gamma(t) = (a_1,\dots,a_k -t,\dots,a_n)$ in $ \mathbb{R}^n$. Then we have
$$ \nabla_{d/dt} e_j = - \sum_i \Gamma_k^{ij} e_i $$
This implies that a section $f(t)$ along $ \gamma $ is paralell when
$$ f' = (\Gamma_k^{pq}) f$$
Now let $\widetilde{\nabla}$ be the connection induced from our paralell sections. Then
$$ (\widetilde{\nabla}_{\partial / \partial x_k} e_j)_{(a_i)} = \lim_{t \to 0} \frac{T(e_j) - e_j}{t} = \sum_i \Gamma_{k}^{ij}(a_1,\dots,a_n) e_i $$
Therefore $ \widetilde{\nabla} = \nabla$. I think this answers your question, but I want to say a little more. The parallel transport of $e_j$ in the negative $x_k$ direction is
$$ e_j + (\Gamma_k^{ij})e_j t + O(t^2) $$
Therefore, the paralell transport of the frame $ I = \{ e_1,\dots,e_d \} $ in the negative $x_k$ direction is
$$ I + (\Gamma_k^{ij}) t + O(t^2) $$
Therefore the splitting $ \sigma : \pi^* TM \to T {\rm Fr} \, E $ is defined by
$$ (\partial / \partial x_k)_{(m,I)} \mapsto (\partial / \partial x^k, - \Gamma_k^{ij}) $$
so the connection form $ \omega$ is defined by
$$ \omega_{(m,I)} (\partial / \partial x_k, a^{ij}) = a^{ij} + \Gamma^{ij}_k $$
Locally, the connection form is defined by a $ \mathfrak{gl}_n$-valued 1-form on $M$. This computation demonstrates that this form is exactly the Christoffel symbols $ \Gamma^{ij}_k $.
