You have $ \bigl\vert\vert f_{k+l}(x)\vert-\vert f_{k}(x)\vert\bigr\vert\le \vert f_{k+l}(x)-f_{k}(x)\vert $ and thus $$ \Vert\vert f_{k+l}\vert-\vert f_{k}\vert\Vert_{W^{1,1}} \le \Vert f_{k+l}- f_{k}\Vert_{L^{1}}+\Vert \nabla{\vert f_{k+l}\vert} -\nabla{\vert f_{k}\vert}\Vert_{L^{1}}. $$ Now, let us calculate $\nabla \vert f\vert$, say for $f\in C^1$. We claim that we have $$ \nabla \vert f\vert=S(f) \nabla f, \quad S(f)=\mathbf 1(f>0)-\mathbf 1(f<0). \tag 1$$ Note that the function $S(f)$ is bounded measurable and that the product makes sense, even if you have only $\nabla f\in L^1$. Let us prove our claim: with brackets of duality and $\phi\in C^\infty_c$, we have \begin{multline} \langle \nabla \vert f\vert, \phi\rangle=-\int\vert f\vert \nabla \phi dx=-\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{1/2} \nabla \phi dx= \lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f) \phi dx \\ =\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f) \phi dx =\int S(f) (\nabla f) \phi dx, \end{multline} proving (1), using Lebesgue's Dominated Convergence Theorem. I feel a bit uncomfortable to extend this formula to $f\in W^{1,1}$, but it might be a first step.

You have $ \bigl\vert\vert f_{k+l}(x)\vert-\vert f_{k}(x)\vert\bigr\vert\le \vert f_{k+l}(x)-f_{k}(x)\vert $ and thus $$ \Vert\vert f_{k+l}\vert-\vert f_{k}\vert\Vert_{W^{1,1}} \le \Vert f_{k+l}- f_{k}\Vert_{L^{1}}+\Vert \nabla{\vert f_{k+l}\vert} -\nabla{\vert f_{k}\vert}\Vert_{L^{1}}. $$ Now, let us calculate $\nabla \vert f\vert$, say for $f\in C^1$. We claim that we have $$ \nabla \vert f\vert=S(f) \nabla f, \quad S(f)=\mathbf 1(f>0)-\mathbf 1(f<0). \label{1}\tag{1}$$ Note that the function $S(f)$ is bounded measurable and that the product makes sense, even if you have only $\nabla f\in L^1$. Let us prove our claim: with brackets of duality and $\phi\in C^\infty_c$, we have \begin{multline} \langle \nabla \vert f\vert, \phi\rangle=-\int\vert f\vert \nabla \phi dx=-\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{1/2} \nabla \phi dx= \lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f) \phi dx \\ =\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f) \phi dx =\int S(f) (\nabla f) \phi dx, \end{multline} proving \eqref{1}, using Lebesgue's Dominated Convergence Theorem. I feel a bit uncomfortable to extend this formula to $f\in W^{1,1}$, but it might be a first step.