First, we can observe that your integral depends solely on the behavior of $f$ on the interval $[0, 1]$. Its values outside that region do not affect the expression. So we may multiply by the cutoff function $\chi_{[0, 1]}$.

Thus we may consider this problem for elements of $L^2(\mathbb{R})$ with compact support.

We can look at the Hardy-Littlewood-Sobolev theorem on fractional integration (or more accurately, its proof). The argument found in Remark 2 here, partitioning into dyadic shells, shows that for $x > 0$, $$\left|\int_0^x \frac{f(y)}{|x - y|^{1/2}} \, dy \right| \leq \int_0^x \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq \int_{B(x, x)} \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq C x^{1/2} M f(x),$$ where $M f$ denotes the Hardy-Littlewood maximal function and $C$ is an absolute constant.

In your case, since we are integrating over $x \in [\epsilon, 1]$, we can bound this solely by a multiple of $M f$, and then the strong-type
Hardy-Littlewood $L^p$ estimate gives (as a *very* rough upper bound), $$\left\|\int_0^x f(y) |x - y|^{-1/2} \, dy \right\|_{L^2_x[0, 1]} \leq C \|M f(x)\|_{L^2_x} \leq C' \|f\|_{L^2} = C' \|f\|_{L^2[0, 1]},$$ invoking the support restriction on $f$.

So your integral is bounded above by the quantity $K \int_{0}^{1} |f(x)|^2 \, dx$, for some dimensional constant $K$. (And this also indicates that you don't need to take $\epsilon > 0$; you can directly integrate over $[0, 1]$.) You specified $f$ is locally $L^2$, so this quantity is well-defined and non-infinite.

Thus, for any $f \in L^2_{\text{loc}}$, you can guarantee that your integral will be finite.