Let $\mathbb{O}$ is the sheaf of germs of holomorphic functions on $\mathbb{C}$, $p: \mathbb{O} \rightarrow \mathbb{C}$ defined by sending the germ $f_a$ to $a \in \mathbb{C}$.
Given a path $\gamma: [0,1] \rightarrow \mathbb{C}$, $f_a$ a germ such that $\gamma(0) = a$, if there is a a lift $\gamma':[0,1] \rightarrow \mathbb{O}$ satisfying $\gamma' \circ p = \gamma$, then we say we can analytically continue $f$ along the path $\gamma$.
Now I can sort of see why this is a way of analytically continue $f$, since for each point $x$ in $Im(\gamma)$, because of the lift, we can find $g_x$ a germ in $Im(\gamma')$, such that $p(g_x) = x$. $g_x$ has some representative $(g,V)$, and we can extend $f$ domain to include $V$, and $f=g$ on $U \bigcap V$ ($U$ domain of $f$) (this is the step I am most worried about, since I am not sure why the 2 germs should agree on the intersection)
Now the above is my rough intuition about how this analytically continued business works, but I would like to see a proof that shows that the above conditions are enough to give us a new analytic function that analytically continue $f$.