Suppose $M$ is a manifold, and $E$ a vector bundle over $M$ equipped with a connection $\nabla $. If $F$ is the frame bundle of $E$, is there an explicit construnction of a connection on $F$ associated with $\nabla$ such that in this way connections on $E$ and $F$ are $1$-$1$ correspondent?
Edit for the bounty:
I really need an answer to this question, and as it was already posted I think that putting a bounty on it is the most sensible way to go.
To rephrase the question in my own terms: Let $M$ be a smooth $n$-manifold. We can associate the following principal $GL(n)$-bundle to it: $$F = \{(m,\theta)|m\in M, \theta:\mathbb{R}^n\to T_mM\mathrm{\ lin.\ isom.}\}$$ with right action given by $(m,\theta)g = (m,\theta g)$. Its tangent space is defined (as for any other manifold) as a quotient of the space of paths on $F$. In order to get a more concrete representation, we need a way to differentiate "paths of frames," but as such paths can be seen as tuples of paths of vectors on $M$, it is enough to specify a connection $\nabla$ on $M$ to obtain the identification $$T_{(m,\theta)}F \cong \{(\hat{m},\hat{\theta})|\hat{m}\in T_mM,\hat{\theta}:\mathbb{R}^n\to T_mM\}$$ where we identify the equivalence class of paths $[\gamma(t),\theta(t)]$ with $(\dot{\gamma}(0),(\nabla_{\dot{\gamma}}\theta)(0))$. This gives us a map $$\{\mathrm{connections\ on\ }M\}\longrightarrow\{\mathrm{principal\ connections\ on\ F}\}$$ mapping $\nabla$ to $A([\gamma,\theta]) = \theta^{-1}\nabla_{\dot{\gamma}}\theta\in\mathfrak{gl}(n)$.
I believe there should be a way to invert this map (maybe only on a subset of the principal connections, though) but I cannot see how. Does anyone have an idea or a solution?
Remark 1: My question is in fact a special case of the original question on vector bundles, namely if we take $E=TM$.
Remark 2: I took a look at Taubes' book, as suggested in the answers, but I didn't find what I need (or maybe I found it, but wasn't smart enough to realize it).