I think that I misunderstood the notation in Question 1. The bold statement is incorrect. Consider the function $\tau(x) = 2x$ on $\mathbb{R}$ and the function $k(x) : = x$ on $\mathbb{R}$. Then $S_t = \{t/2\}$. So, at time $t$, the value of $k(x)$ on $S_t$ is $t/2$. So, the rate of change is $1/2$. On the other hand $\xi$ would be $\frac{\partial}{\partial x}$, so $\mathscr{L}_\xi k = 1$. 

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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)