Here's a rough idea of what is going on. And I'm going to restrict my comments to the torsion free case which simplifies the language (and, I think, is implicit in your question anyway).
Let $\mathcal H$ be the convex hull of $\Lambda(\Gamma)$, and consider the quotient $\mathcal H / \Gamma$ which is embedded in $M$. Outside of some exceptional situations (e.g. Fuchsian groups), $\mathcal H / \Gamma$ is homeomorphic to $M$, and the difference between them, by which I mean the closure $\overline{M - \mathcal H / \Gamma}$, is homeomorphic to $\partial M \times [0,1]$. So, in particular, we have a rather natural homeomorphism $S \approx \partial \mathcal H / \Gamma$. (One can even get around the exceptions by replacing $\mathcal H$ with its $\epsilon$-neighborhood, as is explained in work of Thurston that one can find developed in work of Epstein and Marden; but I will stick with this simpler notation.)
Of course, under the assumption that $S$ is not compact, i.e. the number of punctures $n_i$ is nonzero for some $i$, it follows that $\mathcal H / \Gamma$ is not compact. So $\mathcal H / \Gamma$ is the "convex core" of $M$, but not yet its compact core.
To get the actual compact core of $M$, you now look at the "cusps" of $\mathcal H / \Gamma$. The compact core itself is obtained by cutting off each cusp and adding a new boundary portion associated to that cusp. Cusps are of two types: $\mathbb Z \oplus \mathbb Z$ cusps; and $\mathbb Z$ cusps. When you cut off a $\mathbb Z \oplus \mathbb Z$ cusp of $\mathcal H / \Gamma$, you just add a new boundary component homeomorphic to the torus. But when you cut of a $\mathbb Z$ cusp of $\mathcal H / \Gamma$, what you add is a cylinder that connects up two circle cusps of the Riemann surface $S = \partial \mathcal H / \Gamma$.
So to summarize, via the inclusions $C \subset \mathcal H / \Gamma \approx M$, one can think of $S$ as embedded in $\partial C$, in such a way that the complement $\partial C - S$ has components in one-to-one correspondence with the cusps of $\mathcal H / \Gamma$: a torus component for each $\mathbb Z \oplus \mathbb Z$ cusp; and an annulus component for each $\mathcal Z$ cusp.
I am assuming that by complete you mean all integral curves of a vector field $X$ can be extended to all of $\mathbb{R}$. I also haven't thought to much about this specific case before, so if I say something incorrect, my apologies.
You are correct that the frame bundle of a Lie group $G$ has a global section. In particular, this is because a basis for the Lie algebra determines a global frame, hence if $(T_1,\dots,T_n)\in T_eG\cong \mathfrak{g}$ is a basis for the Lie algebra, then the global section is given by:
$$s(g)=(L_{g*}T_1,\dots, L_{g*}T_n)$$
as at each point $g\in G$, this is a frame for $T_gG$. This gives us a global trivialization of the frame bundle for $G$, hence:
$$\mathfrak{F}(G)\cong G\times GL_n(\mathbb{R})$$
which is itself a Lie group so every vector field is complete, including horizontal lifts.
Ok so for a general principal $G$ bundle $\pi:P\rightarrow M$ how do we obtain a horizontal lift a vector field on $M$? Well we have that any connection one form determines a horizontal sub bundle $H \subset TP$, and it follows that $D_p\pi:H_p\rightarrow T_{\pi(p)}M$ is then an isomorphism for all $p\in P$. We thus fix a connection one form on $P$.
Given $X\in \mathfrak{X}(M)$ we thus define a $X^H\in\mathfrak{X}(P)$ pointwise by:
$$X^H_p=(D_p\pi)^{-1}(X_{\pi(p)})$$
We need to show that this is smooth, so let $\phi$ be a bundle chart for an open neighborhood $U$ of $\pi(p)=x$. Let $\phi(p)=(x,h)$, then $D_p\phi:T_gP\rightarrow T_xM\oplus T_gG$ is an isomorphism. Let $\pi_U:U\times G$ be the projection onto $U$, and define a smooth section of the pullback bundle $\pi_U^*TU$ by:
\begin{align*}
Z:(U\times G)&\longrightarrow \pi_U^*TU\\
(x,g)&\longmapsto ((x,g),X_x)
\end{align*}
which is smooth because $X$ is smooth, and takes values in $\pi^*_U TU\subset \pi^*_UTU\oplus \pi^*_GTG\cong T(U\times G)$, so $Z$ is actually a smooth section of $T(U\times G)$. It is then easy to see that:
\begin{align*}
D_{(x,g)}\pi_U(Z_{(x,g)})=X_x
\end{align*}
Since $\phi$ is a diffeomorphism, it follows that there exists a unique vector field $Y\in\mathfrak{X}(P_U)$ such that $\phi_*(Y)=Z$, hence for all $p\in P_U$:
\begin{align*}
D_p\phi(Y)=Z_{\phi(p)}
\end{align*}
So:
\begin{align*}
D_{\phi(p)}\pi_U\circ D_p\phi(Y)=X_{\pi(p)}
\end{align*}
however:
\begin{align*}
D_{\phi(p)}\pi_U\circ D_p\phi=D_p(\pi_U\circ \phi)=D_p\pi
\end{align*}
so:
\begin{align*}
D_p\pi(Y_p)=X_{\pi(p)}
\end{align*}
Since $X^H$ is uniquely determined, it follows that $X^H$ is locally the horizontal component of $Y$, and thus $X^H$ must be smooth. It follows that $X^H$ is the unique horizontal lift of $X$.
Regarding your question pertaining Lie groups, perhaps you can clarify what you mean specifically, because as I see it now, there's not a clear way to obtain the horizontal lift in terms of a global section. Indeed, sections determine bundle charts/local trivializations, but these bundle charts don't respect the horizontal lift (the above proof demonstrates this clearly) as that is determined by a connection one form. You could choose the flat connection, which exists in this case because the bundle is trivial, but in general it won't be a metric connection, and so can't be the Levi Civita connection.
If it helps at all, you can write the connection one form with on $G\times GL_n(\mathbb{R}) as:
$$(\phi^{-1*}\omega)_{(g,A)}=A^{-1}\pi_G^*\omega_{s(g)} A+\pi_{GL_n(\mathbb{R})}^*(A^{-1}dA)$$
where $\omega_s=s^*\omega$, $A\in GL_n(\mathbb{R})$, and $A^{-1}dA$ is the Maurer Cartan form on $GL_n(\mathbb{R})$.
Best Answer
Here is how the product rule (always) arises in such situations. Let $\xi = fe$. Then (differentiating in the direction of the vector $e = d/dx$ is understood) $$\nabla \xi = \nabla(fe) = f'(x)e + f\nabla e = \big(f'(x)+f(x)\Gamma(x)\big)e.$$ So $\xi$ is parallel if and only if $f'(x)+f(x)\Gamma(x)=0$, which means $f'(x)/f(x) = -\Gamma(x)$. Taking $f(0)=1$, this gives us $$f(x) = f(0)e^{-\int_0^x \Gamma(y)\,dy} = e^{-G(x)}.$$ I presume the text you are reading will discuss the relevance of $\phi$.