I'll abuse notation and write $u(x) \in L^1(\mathbb R)$ for an expression $u(x)$ instead of writing the more correct $x \mapsto u(x) \in L^1(\mathbb R)$
Suppose that $\hat f(\xi) \in L^1(\mathbb R)$. Then also $\hat f(\xi) e^{i \xi x} \in L^1(\mathbb R)$.
By the Fourier inversion formula,
$$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi$$
We want to show that for any $x_0 \in \mathbb R$ we have $\lim_{x \to x_0} f(x) = f(x_0)$.
It's enough to show that $\lim_{x_n \to x_0} f(x_n) = f(x_0)$ for any sequence $x_n \to x_0$.
Let $\hat f_n(\xi) = \hat f(\xi) e^{i \xi x_n}$ and $g(\xi) = |\hat f(\xi)|$. Since $e^{i \xi x}$ is continuous in $x$, we have pointwise convergence, i.e. $\hat f_n(\xi) \to \hat f(\xi)$ for every $\xi \in \mathbb R$. Also, for all $n$, $|\hat f_n(\xi)| \leq g(\xi)$ where $g(\xi) = |\hat f(\xi)| \in L^1(\mathbb R)$.
By the dominated convergence theorem then $\int \lim_{n \to \infty} \hat f_n(\xi) d\xi = \lim_{n \to \infty} \int \hat f_n(\xi) d\xi$, i.e.
$$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi
= \frac{1}{2\pi} \lim_{n \to \infty} \int \hat f_n(\xi) d\xi
= \lim_{n \to \infty} \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x_n} d\xi
= \lim_{n \to \infty} f(x_n)$$
Thus, $f$ is continuous.
Suppose $f(x)= 0$ for $x\ne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $\mathbb R.$ Since $f=0$ a.e., we have $F[f]\equiv 0,$ and hence $F[F[f]](x)\equiv 0.$ Thus $F[F[f]](x)$ does not equal $2\pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:\mathbb R\to \mathbb R$ is continuous on $(-\infty,0)\cup(0,\infty)$ and $f\in L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]\in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -\frac{F[F[f])(-x)}{2\pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-\infty,0).$ But two continuous functions that agree a.e. on $(-\infty,0)$ actually agree everywhere on $(-\infty,0).$ (Nice exercise) The same holds on $(0,\infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
Best Answer
Yes. In fact one derivative is enough.
The Fourier transform of $f'$ is $itF(t)$ (or something like that, depending on how your definition of the Fourier transform is normalized). Since $f'\in L^1$ this shows that $tF(t)$ is bounded. So $$\int_{|t|>1}|F(t)|^2\,dt\le c\int_{|t|>1}\frac{dt}{t^2}<\infty.$$ We certainly have $\int_{|t|\le1}|F(t)|^2\,dt<\infty$ since $F$ is bounded. So $F\in L^2$ and hence $f\in L^2$.
Bonus We've proved this:
Theorem If $f\in L^1(\Bbb R)$ is absolutely continuous then $$||f||_2\le c(||f||_1+||f'||_1).$$That can't be "right", that is it can't be the whole truth, because the two sides transform differently under dilations. If we use the argument above but split the integral of $|F|^2$ at $|t|=A$ instead of $|t|=1$ we get this: $$||f||_2\le c(A^{1/2}||f||_1+A^{-1/2}||f'||_1).$$Now let $A=||f'||_1/||f||_1$:
Better Theorem If $f\in L^1(\Bbb R)$ is absolutely continuous then $$||f||_2\le c||f||_1^{1/2}||f'||_1^{1/2}.$$