Do there exist functions $f: \mathbb{R} \to \mathbb{R}$ with the following properties?
1: $f$ is at least $C^3$ differentiable.
2: $f$ and at least its first three derivatives vanish at infinity.
3: $0<\int_{-\infty}^\infty f(x)^2dx < \infty$
4: $\operatorname{sign}(f) = \operatorname{sign}(f'')$ over the entire domain $\mathbb{R}$.
No, there is no such function $f$.
Note that $f$ is continuous and vanishes at $\pm\infty$ (you only said "at infinity", but the integral condition implies vanishing at both positive and negative infinity). Therefore, $|f|$ must achieve a maximum at some point $x \in \mathbb{R}$. By replacing $f$ with $-f$ as necessary, we may assume without loss of generality that $f(x) > 0$, and so, $f(y) \le f(x)$ for all $y \in \mathbb{R}$.
Of course, $f''(x) > 0$ too. Since $f''$ is continuous, there must exist an $\varepsilon > 0$ such that $f$ is strictly convex on $\mathcal{U} = (x - \varepsilon, x + \varepsilon)$. But then, $$f(x) = f\left(\frac{x + \frac{\varepsilon}{2}}{2} + \frac{x - \frac{\varepsilon}{2}}{2}\right) < \frac{1}{2}f\left(x + \frac{\varepsilon}{2}\right) + \frac{1}{2}f\left(x - \frac{\varepsilon}{2}\right) \le \frac{1}{2}f(x) + \frac{1}{2}f(x) = f(x),$$ which is a contradiction.