If $(f_n)\to0$ in $L^2$ and the derivatives converge in $L^2$, must the limit be 0?

Due to your assumptions $(f_n)$ converges in $L^2(\Omega)$, and $(f_n')$ is a Cauchy sequence in $L^2(\Omega)$ hence convergent. That is, $f_n \to f$ and $f_n'\to g$ in $L^2(\Omega)$.

Let $\phi\in C_c^\infty(\Omega)$ be given. Passing in the equation $$ \int_\Omega f_n \phi' = -\int_\Omega f_n' \phi $$ to the limit yields $f'=g$ in the sense of weak derivatives. Then it is not hard to conclude $f_n\to f$ in $H^1(\Omega)$.

[In the question $f=0$, then also $f'=g=0$ follows.]

We are given $f_n \to 0$ in $L^2 = L^2[0,1].$ We are also given $(f_n')$ is Cauchy in $L^2.$ Since $L^2$ is complete, $f_n'$ converges in $L^2$ to some $g\in L^2.$ (Let's note for later use that this implies $f_n' \to g$ in $L^1.$)

You are asking if $g=0$ in $L^2.$ Suppose this fails. Then $g\ne 0$ on some set of positive measure. WLOG, $g>0$ on some $E\subset [0,1]$ with $m(E)>0.$

Apply the Lebesgue differentiation theorem to $g$ on $E:$ For a.e. $a\in E,$ we have

$$\tag 1 \lim_{h\to 0^+} \frac{1}{h}\int_a^{a+h} g(x)\,dx = g(a)>0.$$

Now $f_n \to 0$ in $L^2$ implies there is a subsequence $f_{n_k}$ that converges pointwise to $0$ a.e. It follows that there exists $a_0\in E$ such that i) $(1)$ holds with $a=a_0$ and ii) $f_{n_k}(a_0) \to 0.$

For this $a_0$ there then exists $\delta > 0$ such that

$$\tag 2 \int_{a_0}^{a_0+h} g(x)\,dx > (g(a_0)/2)h\,\,\text { for } 0<h<\delta.$$

Again using $f_{n_k} \to 0$ a.e., we have $f_{n_k}(a_0+h_0) \to 0$ for some $h_0\in (0,\delta).$

We have arrived at two incompatible results: By $(2)$ we have

$$\tag 3 \int_{a_0}^{a_0+h_0}f_{n_k}'(x) \,dx > (g(a_0))/2)h_0$$

for large $k,$ but by smoothness, $(3)$ equals $f_{n_k}(a_0+h_0)-f_{n_k}(a_0),$ and both of these terms $\to 0$ as $k\to \infty.$

This contradiction shows $g=0$ in $L^2,$ and we're done.

While this question does start to hint at the utility of Sobolev spaces, you do not really require that machinery here. First, recall that convergence in $L^2$ implies convergence pointwise almost everywhere along a subsequence. So there exists $(n_k)$ such that $f_{n_k}(x)\to0$ for all $x\in A$, where $A\subset[0,1]$ has full measure.

Next, we know that $(f_n')$ is a Cauchy sequence in $L^2(0,1)$, so $f_n'\to g$ in $L^2$ for some $g\in L^2(0,1)$. For any $a,b\in(0,1)$, we have

$$\left|\int_a^bf_n'(t)dt-\int_a^bg(t)dt\right|\le\int_0^1|f_n'-g|dt\le\left(\int_a^1|f_n'-g|^2dt\right)^{1/2}\to0$$

as $n\to\infty$. Since $f_n(b)-f_n(a)=\int_a^bf_n'(t)dt$, choosing $a,b\in A$ and $n=n_k$ implies

$$\left|\int_a^bg(t)dt\right|\le|f_{n_k}(b)-f_{n_k}(a)|+\left|\int_a^bf_n'(t)dt-\int_a^bg(t)dt\right|\to0$$

as $k\to\infty$. From here it is straightforward to verify that $\int_Bg(t)dt=0$ for any Borel $B\subset[0,1]$, and so $g=0$ almost everywhere. This completes the proof.