We have $$d(t):=\int dx\Big(\frac{f(x+t)-f(x)}{t}-f'(x)\Big) =\int dx\int_0^1 ds\,(f'(x+st)-f'(x)),$$ whence $$|d(t)|\le\int_0^1 ds\, J_t(s),$$ where $$J_t(s):=\int dx\,|f'(x+st)-f'(x)|\underset{t\to0}\longrightarrow0 \tag{10}\label{10}$$ for each $s$, since $f'\in L^1$ (see e.g. this answer). Also, $$|J_t(s)|\le\int dx\,(|f'(x+st)+|f'(x)|)=2\|f'\|_1<\infty.$$
So, by dominated convergence, $d(t)\underset{t\to0}\longrightarrow0$.
Instead of dominated convergence, one can use the fact that the convergence in \eqref{10} is uniform in $s\in[0,1]$.