I'm slightly confused by your notation. I think that you have switched the unit normal and conormal. In particular in the physicist GR notation, the coefficients of a vector field should have _raised_ indices, which you can remember by
$$
V = V^i \frac{\partial}{\partial x^i}
$$
so you're summing over one "raised" $i$ and one "lowered" $i$. So, I think that you're mixing up vector fields and 1-forms. 

But $\xi^a$ is the unit normal vector field to $\{\tau = c\}$ as far as I can tell, so what you've written in "Question 1" is correct.

As for why Question 1 holds, perhaps you should review the definition of the Lie derivative. The correct  "definition" of the Lie derivative is the one given <a href="http://en.wikipedia.org/wiki/Lie_derivative#Lie_derivative_of_tensor_fields">here</a>, namely if $T$ is a tensor and $\xi$ a vector field generating the path of diffeomorphisms $\varphi_t$ then
$$
\mathcal{L}_\xi T = \frac{d}{dt}\Big|_{t=0} \varphi_t^* T.
$$
So, in your case $\xi$ is a vector field on $\Sigma$ and along flowlines of $\xi$, $\tau$ is increasing at a constant unit rate (do you see why?). Now, you should think through why this answers Question 1.

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As for question 2, I am not going to do the computation for you, but let me explain in modern notation how to understand. I will write $S_c:=\{\tau=t\}$ and $H$ for the mean curvature of $S_t$. Consider the quantity
$$
\int_{S_t} H^2 d\mu_t
$$
We let $u = |\nabla \tau|^{-1}$ as above. This is often called the _lapse function_. Now, how do we compute the rate of change of this integral?
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \frac{d}{dt} H^2 d\mu_t + \int_{S_t} H^2 \frac{d}{dt} d\mu_t
$$
(here, I am thinking of integrating over a fixed, abstract sphere, where the function $H$ and measure $\mu_t$ are time dependent).

What is the first term? To differentiate $H$, we must use the second variation formula,
giving
$$
\frac{d}{dt} H = -\Delta_{S_t} u -(Ric(\xi,\xi)+\Vert h \Vert^2)u.
$$
Here, $h$ is the second fundamental form. To differentiate the second term, one should use the first variation formula, giving
$$
\frac{d}{dt} d\mu_t = uH d\mu_t.
$$
Thus, putting these together yields
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \left(-2H \Delta u - 2H(Ric(\xi,\xi) + \Vert h\Vert^2)u + uH^3 \right) d\mu_t
$$
Now, using the Gauss equations, we have that 
$$
2(Ric(\xi,\xi) + \Vert h\Vert^2) = R-\mathcal{R}+\Vert h \Vert^2 + H^2 
$$
Inserting this into the above equation yields
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \left(-2H \Delta u - uHR +uH\mathcal{R} -uH\Vert h\Vert^2  \right) d\mu_t
$$
This is exactly your equation.

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I'll mention that an apt choice is $u=\frac{1}{H}$. This yields the so-called inverse mean curvature flow. It would be very instructive for you to plug this in and try to find a nice differential inequality for what you call $C(\tau)$ assuming that $R\geq0$. 

Of course, Geroch is unconcerned with the existence of such a function $\tau$. This turns out to be a very serious problem, and was only recently solved in the beautiful work of Huisken--Ilmanen in their proof of the Penrose inequality: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1090349447. In particular, the computation I have just done is contained in this article for the special case of inverse mean curvature flow (as this is the only case that is really important)