When viewing Sobolev spaces as completions, how does the notion of weak derivative arise?

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: $$ \int \nabla f \varphi = - \int f \operatorname{div} \varphi \quad \text{for } f \in W^{1,p}(\Omega). $$