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:
$$ \int \nabla f \varphi = - \int f \operatorname{div} \varphi \quad \text{for } f \in W^{1,p}(\Omega). $$
Best Answer
I cannot explain all the features, but the defining property of a compactly supported function $f$ defined on an open set $\Omega$ is that there exists a compact set $K$ in $\Omega$ such that $f(x)=0$ if $x\not\in K$. This is useful since for instance when you do integration by parts on $f$, the boundary terms vanish. If $g$ is any function, you can get a compactly supported function $fg$ by multiplying it by $f$. This is called a cut-off process.
In the PDE context, cut-off functions are usually smooth, this means that, as mentioned in the comment, all derivatives of $f$ will be uniformly continuous.