Let $P$ be a principal $G$-bundle, $\rho:G\to GL(V)$ a finite dimensional representation of $G$, $E = P \times_G V$ the associated vector bundle. To any principal connection $\Phi$ on $P$ is associated an induced linear connection $\bar \Phi$ on $E$. Conversely, any linear connection on a vector bundle $E$ is induced from a unique principal connection on the linear frame bundle $GL(\mathbb R^n, E)$ of $E$.
You can find all the details here: Topics in Differential Geometry - P. W. Michor.
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 $.
Best Answer
For 1, given any free action of a compact Lie group $G$ on a manifold $M$ (or a free proper action of a noncompact Lie group), the orbit space $M/G$ naturally has the structure of a smooth manifold such that the projection $\pi:M\rightarrow M/G$ is a smooth submersion. (Free means the only group element which fixes at least one point is the identity).
It turns out that $\pi$ is actually a $G$-principal fiber bundle. So, if you care about group actions on manifolds, principal bundles arise naturally.
For 3, one of the main uses of the frame bundle I know of is the following: Suppose a compact Lie group $G$ acts effectively on a Riemannian manifold $M$. (Effective means the only group element which fixes all points is the identity). Since the action is not free, the orbit space $M/G$ isn't a manifold in any kind of natural way (though it is still not so bad as a topological space!).
On the other hand, the action induces an action on the tangent bundle $TM$ (which still might not be free), and induces and action on the frame bundle $FM$. This induced action on the frame bundle is free, so the quotient $FM/G$ is a manifold, so all the tools of differential geometry can be used to study $FM/G$, which in turn can give information about $M$.