Given a fiber bundle $\pi:E\to M$, a curve $\gamma:[0,1]\to M$, and a point $p \in \pi^{-1}(\gamma(0))$, a connection on the bundle allows us to uniquely lift $\gamma$ to a horizontal curve in E through $p$. In almost all situations I have encountered, the horizontal lift does not depend on the orientation of $\gamma$. To be precise, the two curves $t\to \gamma(t)$ and $t\to\gamma(1-t)$ have the same horizontal lift through $p$.

I have a fiber bundle for which I would like to have a type of parallel transport which depends on which direction one is moving in the base. So my question is: what is the best way to formulate a connection which is orientation dependent, and so enables this type of parallel transport?

fixing a point $x$ in fibre of $\gamma(0)$, there exists a curve that starts at $x$. So, lift hasstarting pointas $x$. Suppose you choose $t\mapsto \gamma(1-t)$, you fix a point$y$ in fibre of $\gamma(1)=\gamma(1-0)$, you get a lift whose starting point is $y$. Thisdoes not sayhorizantal lift of $\gamma(t)$ and $\gamma(1-t)$ are same if you are looking fromorientationperspective. What is that I am misunderstanding in your question? $\endgroup$notthe same (their images are), so the usual lift is already orientation dependent. $\endgroup$