Here $D^h u = (u(\cdot+h)-u(\cdot))/|h|$, if I recall correctly. The issue is that while the $L^1$ norm of $D^hu$ controls the total variation of $u$, it does not control how concentrated this variation can be, e.g., it does not prevent the variation from being supported on a set of measure zero. The uniform boundedness of $\|D^h u\|_{L^1}$ yields $u\in BV$, which is a larger space than $W^{1,1}$. (In contrast, the $L^p$ norm of function $f$ for $p>1$ does control the degree of concentration of $f$, via Hölder's inequality $|\int_E f|\le \|f\|_p |E|^{1/p'}$.)
For a concrete example, take $u$ to be any discontinuous function of bounded variation, such as the characteristic function of a ball $B$ contained in $V$. The difference $\chi_B(\cdot+h)-\chi_B(\cdot)$ has $L^1$ norm of magnitude $|h|$. The computation is particularly easy in one dimension.