If we take $f$ to be a smooth function, then how does it follow that we can write $f^{\epsilon}(x)-f(x) = \int_{B(0,1)}\eta(y)(f(x-\epsilon y)-f(x))dy$ where $f^{\epsilon} := \eta_{\epsilon}\ast f$ where $(\epsilon>0)$ and
$\eta_{\epsilon}$ denotes the usual mollifier.
First part (by Tim kinsella): $$RHS=\int \eta (y)f(x-\epsilon y) dy- f(x) \int \eta (y) dy = \int \frac{\eta(y/ \epsilon)}{\epsilon} f(x-y) dy - f(x) = LHS$$ Answering follow-up question: if $V$ is the entire space, then for any $t,y$ $$ \int_{V}|Df(x-\epsilon t y)|dx = \int_{V}|Df(x )|\,dx $$ Subsequent integration in $t$ and $y$ yields $1$.