In a metric space $(X,d)$, you may look at the length functional as a FUNCTIONAL defined over the space of Lipschitz functions from $[0,1]$ to $(X,d)$. One of the natural notions of convergence of functionals is the gamma-convergence.
For a sequence of metric spaces $(X_{n},d_{n})$ which converges (in the Gromov-Hausdorff distance) to a metric space $(X,d)$, we know that there is a metric space $(Y,D)$, a sequence of isometric embeddings $F_{n}: (X,d_{n}) \rightarrow (Y,D)$ and another isometric embedding $F:(X,d) \rightarrow (Y,D)$, such that the sequence $F_{n}(X_{n})$ Hausdorff converges to $F(X)$.
A natural question would be: is there a sequence of isometries $f_{n}: (X,d_{n}) \rightarrow (Y,D_{n})$ and an isometry $f:(X,d) \rightarrow (Y,D)$, such that
$D_{n}$ and $D$ generate the same uniformity on $Y$,
$D_{n}$ converges uniformly (on compact sets, on bounded sets, etc, pick your choice) to $D$,
pick the topology of uniform convergence on the space $\mathcal{C}([0,1],Y)$.
Then the sequence of length functionals associated to $(f_{n}(X_{n}), D_{n})$ gamma converges to the length functional of the limit $(f(X), D)$.
If true then any sequence of $D_{n}$ geodesics (length minimizers) converges to a geodesic of $D$, for example.
This is true in a Riemannian manifold (seen as the metric space $(X,d)$), if you take $(X_{n},d_{n})$ as $(B(x,1/n), \frac{1}{n}d)$, which GH converges to the (unit ball in the) tangent space at $x \in X$, for every point $x$. (This follows from results from the paper G. Buttazzo, L. De Pascale, I. Fragala,
Topological equivalence of some variational problems involving distances, Discrete Contin. Dynam. Systems 7 (2001), no. 2, 247-258).
It is also true for sub-riemannian manifolds, see the paper M. Buliga, A characterization of sub-riemannian spaces as length dilation structures constructed via coherent projections, Commun. Math. Anal. 11 (2011), No. 2, pp. 70-111, arxiv link.
All this is related to the metric characterization of riemannian (and sub-riemannian) spaces, but in a different way than the great paper I.G. Nikolaev, A metric characterization of riemannian spaces, Siberian Advances in Mathematics, 1999, v. 9, N4, 1-58, MathSciNet link.
Question 1. Yes sure and the same proof should work.
Question 2. The answer is "yes" in $W^{1,p}$ for any $p<\infty$ and "no" in $W^{1,\infty}$.
$W^{1,1}$:
Note that
$$d_{\mathrm{H}}(\Sigma_t,\Sigma_s)<\varepsilon
\ \ \iff\ \
\sup_x\{\,|d_{\Sigma_t}(x)-d_{\Sigma_s}(x)|\,\}<\varepsilon.$$
So it is sufficient to show the following:
If (1) $\sup_x|f_t(x)-f_0(x)|\to 0$ as $t\to 0$,
(2) $|\nabla_xf_t|\le 1$ for any $t,x$ and
(3) $|\nabla_xf_0|=1$ for almost all $x$ then
$$\int\limits_M|\nabla_x f_t-\nabla_x f_0|\cdot dx\to 0.$$
Assume contrary.
Then for some fixed $\varepsilon>0$ and any $t>0$
there is a set $C_t$ of measure $>\varepsilon$ such that $|\nabla_x f_t-\nabla_x f_0|>\varepsilon$ for any $x\in C_t$.
Move (almost) along $\nabla f_0$ and integraite both $df_0$ and $df_1$.
You get a positive lower bound on $\sup |f_t(z)-f_0(z)|$.
The later contradicts that $\sup_x|f_t(x)-f_0(x)|\to 0$.
$W^{1,p}$: All this proves that $f_t$ is continuous in $W^{1,1}$.
Since $\nabla f_t$ is bounded, you get continuity in $W^{1,p}$ for any $p<\infty$.
$W^{1,\infty}$: There is no continuity in $W^{1,\infty}$; this can be seen for $M=\mathbb R$ and $f_t(x)=|x-t|$. (The same works for compact $M$)
Best Answer
I think you may have more restrictive hypotheses in mind, because in general this will be false due to behavior that may be obviously absent in the cases you want to consider. Anyway, here are two counterexamples.
I see $\mathbb S^n$ as the set of $(x_0,\ldots,x_n)\in\mathbb R^{n+1}$ such that $x_0^2+\cdots+x_n^2=1$.
Define $F:(-1,1)\times\mathbb S^n\to\mathbb R^{n+1}$ as $$ (t,x)\mapsto(tx_0,x_1,\ldots,x_n) $$ (I am squashing the sphere on itself like I eat a burger). Then $F_\varepsilon(\mathbb S^n)$ is basically two copies of a disc glued along their boundaries for $\varepsilon$ very small, but $F_0(\mathbb S^n)$ is just one of those discs, so the volume drops at this time. If you want to make this precise, you can find the surface area of the ellipsoid on the dedicated Wikipedia article, and check that for $\alpha$ such that $\mathcal H^2=\alpha\operatorname{Leb^2}$ over $\mathbb R^2$, the Hausdorff volume of $F_\varepsilon(\mathbb S^2)$ is $$ 2\alpha\pi\big(1+\varepsilon^2\log(1/|\varepsilon|)+O(\varepsilon^2)\big) $$ for $\varepsilon\neq0$ small in absolute value, and $F_0(\mathbb S^2)=\alpha\pi$.
Let $G:\mathbb S^n\to\mathbb R^{n+1}$ be a smooth embedding such that $G(x)=(0,x_1,\ldots,x_n)$ for all $x$ with $x_0>0$, and $\rho:\mathbb R\to\mathbb R$ a smooth function with $\mathbf1_{\{0\}} \leq\rho \leq \mathbf1_{(-1/2,1/2)}$; these can be constructed by hand. Define $$ F:(t,x)\mapsto G(x) + t\mathbf1_{x_0>0} \rho(x_1^2+\cdots+x_n^2)\sin(x_1/t^2)e_0 $$ for $t>0$, and $F(t,x)=G(x)$ for $t\leq0$. What is happening is that I am adding ripples on the surface of the sphere, smaller and smaller but also wilder and wilder as $t\to0$.
The continuity of $F$ can be checked sequentially, which becomes easy once one realizes that $\|F_t-F_0\|_\infty\leq |t|$. I claim, and I will not prove, that the volume of $F_\varepsilon(\mathbb S^n)$ is of order $1/\varepsilon$ as $\varepsilon\to0$, $\varepsilon>0$. Clearly the volume cannot be continuous then.
The easiest case $n=1$, which in my opinion already gives a good idea of what is happening, can be treated as follows. For positive $t$, the Hausdorff measure of $F_t(\mathbb S^n)$ is lower bounded by a constant times the length of $$ \big\{(t\rho(y^2)\sin(y/t^2),y),y\in(-1,1)\big\}. $$ Writing $\gamma_t(y)$ for the first coordinate of this point, this length is $$\begin{aligned} \int_{-1}^1\sqrt{1+\gamma_t'(y)^2} \, \mathrm dy &\geq \int_{-1}^1|\gamma_t'(y)| \,\mathrm dy \\ &\geq \int_{-1}^1\Big(\frac1t\rho(y^2) \left|\cos(y/t^2)\right| - 2t |y\rho'(y^2)\sin(y/t^2)| \Big) \, \mathrm dy \\ &\geq t\inf_{|y|<\delta}\rho(y^2)\cdot\int_{-\delta/t^2}^{\delta/t^2} \left|\cos(y)\right| \, \mathrm dy - 4t\|\rho'\|_\infty \\ &\geq \frac\delta t\inf_{|y|<\delta}\rho(y^2) - 4t\|\rho'\|_\infty \\ \end{aligned}$$ for all $\delta>0$ (the fact that the last integral, say over $[-M,M]$, is at least $M$, is true but not obvious; that said, it is clearly asymptotically $4M/\pi$). Fixing $\delta>0$ small enough that the infimum is positive, we get the expected lower bound of order $1/t$. If we wanted to, we could find an upper bound of the same order along the same lines.
My guess, based on very little (basically just the fact that it rules out both examples above), is that this will be true if the function $(t,x)\to F_t(x)$ is jointly Lipschitz and every $F_t$ is injective (since $\mathbb S^n$ is compact, it imposes that it is actually an embedding, for instance). But I think this is a different question altogether.