When viewing Sobolev spaces as completions, how does the notion of weak derivative arise? Suppose instead of defining the Sobolev spaces $W^{k,p}(\Omega), \Omega\subset\Bbb R^n$ as the space of functions whose Sobolev norm (with weak derivatives) is finite, we define it as the completion of the subset of $C^\infty(\Omega)$ functions whose Sobolev norm is finite (call it $C^{k,p}(\Omega)$). By a theorem of topology, these spaces are homeomorphic, since $C^{k,p}(\Omega)$ is dense in both $W^{k,p}(\Omega)$ and $\text{comp}(C^{k,p}(\Omega))$. So while these constructions are equivalent...that's not very clear. 
For instance, it is not clear what $Df$ is supposed to mean, when $f$ is an equivalence class of Cauchy sequences in $C^{k,p}(\Omega)$. (I also don't know how to interpret the equivalent class as a function $f:\Omega\to\Bbb R$ in the first place. I imagine it is an equivalence class of functions that agree a.e. somehow.) In the "standard" viewpoint, we have that $Df$ is just the weak derivative. So how does one interpret what $Df$ means in the completion viewpoint, and can one show that $\int Df\varphi=-\int fD\varphi$, $\forall \varphi\in C^1_0(\Omega)$, just like for weak derivatives?
If this makes sense for $\Bbb R^n$, then I would like to understand it on manifolds. The completion viewpoint seems to be dominant in the geometric analysis literature, but no one explains what $\nabla f$ is actually supposed to be. In Chavel (Eigenvales in Riem. Geo.), we find:

Given a function $f\in L^2(M)$, we say that $Y\in\mathscr L^2(M)$ is a weak derivative of $f$ if 
  $$\int_M\langle Y,X\rangle=-\int_M f\operatorname{div}(X)$$
  for all compactly supported $C^1$ vector fields $X$.

(Here $\mathscr L^2(M)$ are the square integrable vector fields.)
Now, this viewpoint is somewhat different from the usual PDE one since we use compactly supported vector fields...but I suppose this is just a reflection of the fact that $\partial f/\partial x^i$ has no intrinsic meaning on a manifold.
I am either looking for someone to clear up my questions here, or give a good reference on this subject. 
 A: First, you need to separate the concept of completion from the proof of its existence (construction via Cauchy sequences). A completion is just a complete metric space $W^{k,p}$ that contains (an isometric copy of) $C^{k,p}$ as a dense subset. Alternatively, you could use the following universal property to define it:
any uniformly continuous function $f \colon C^{k,p} \to N$ to any complete metric space $N$ has a unique uniformly continuous extension $\bar{f} \colon W^{k,p} \to N$.
For example, the completion of $\mathbb{Q}$ is $\mathbb{R}$, but we usually do not have problems with intepreting elements of $\mathbb{R}$ as numbers. Why? Because we can extend the algebraic operations on $\mathbb{Q}$ to operations on $\mathbb{R}$.
Technically speaking, elements of $W^{k,p}$ in general are not functions, just as it is the case for $L^p$. The reason for this is that there is no meaningful way to define $f(x)$ for chosen $x \in \Omega$. If $f$ is fixed, Lebesgue differentiation theorem states that $f$ is approximately continuous at a.e. $x \in \Omega$ and thus we can make sense of $f(x)$, but the set of admissible points $x$ depends on $f$. It is also worth mentioning that if the product $k \cdot p$ is greater than the dimension of the domain (or if $k$ is equal to the dimension), elements of $W^{k,p}$ can be represented by continuous functions and $f(x)$ makes perfect sense: the evaluation map
$$ W^{k,p} \ni f \mapsto f(x) $$
is continuous for every $x \in \Omega$.
Still, some other useful operations on functions are well-defined on $L^p$, such as integration:
$$ L^p(\Omega) \ni f \mapsto \int_D f, \quad \text{if } D \subseteq \Omega, $$
or multiplication by bounded functions. Of course, this can also be done for $W^{k,p}$.
Now what are weak derivatives? Note that the operation of taking classical gradient
$$ C^{1,p} \ni f \mapsto \nabla f \in L^p $$
is uniformly continuous; remember that $C^{1,p}$ is considered with Sobolev norm. Hence we can extend it continuously to $W^{1,p}$ as its completion, defining the weak gradient $\nabla f$. This should answer one of your questions.
You can easily see that this coincides with the definition you mentioned. Take any $\varphi \in C_c^\infty(\Omega,\mathbb{R}^n)$ and define the linear functional $S_\varphi \colon W^{1,p}(\Omega) \to \mathbb{R}$ by
$$ W^{1,p}(\Omega) \ni f \mapsto \int \nabla f \varphi + \int f \operatorname{div} \varphi. $$
Since the operations of taking weak gradient, multiplying by a bounded function and integrating are well-defined and continuous (in respective spaces and norms), $S_\varphi$ is continuous. On the other hand, $S_\varphi(f) = 0$ for all $f \in C^{1,p}$, which is a dense subset. Hence $S_\varphi \equiv 0$ and we obtain the other definition:
$$ \int \nabla f \varphi = - \int f \operatorname{div} \varphi \quad \text{for } f \in W^{1,p}(\Omega). $$
