Suppose that $f_n$ does not converge uniformly to $0$. Then, there is an $\epsilon>0$ so that for all $N$ there is an $n>N$ and an $x_n$ so that
$$
f_n(x_n)>2\epsilon\tag{1}
$$
Find $\delta>0$ so that $|E|<\delta$ implies
$$
\int_E|f_n^\prime(x)|\,\mathrm{d}x<\epsilon\tag{2}
$$
Suppose that $n$ is such that $(1)$ is true. Let
$$
E_n=\left\{x\in[0,1]:|x-x_n|<\dfrac{\delta}{3}\right\}\tag{3}
$$
Note that $\dfrac\delta3\le|E_n|\le\dfrac{2\delta}{3}<\delta$ (since $x_n$ could be near the boundary of $[0,1]$).
Then $(2)$ says that for $x\in E_n$ we have
$$
|f_n(x)-f_n(x_n)|\le\int_{E_n}|f_n^\prime(t)|\,\mathrm{d}t<\epsilon\tag{4}
$$
The triangle inequality applied to $(1)$ and $(4)$ says that for $x\in E_n$, $f_n(x)>\epsilon$. Thus,
$$
\int_{E_n}f_n(x)\,\mathrm{d}x>\epsilon\frac{\delta}{3}\tag{5}
$$
Since there are arbitrarily large $n$ so that $(1)$ is true, $(5)$ contradicts the assumption that
$$
\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=0\tag{6}
$$