Let $f:\mathbb R\to\mathbb C$ be bounded, continuous, and let
$$\int\left|f\left(x\right)\right|\,dx<\infty.$$
If $x^2f\left(x\right)$ is bounded, then $$\sum_{k\in\mathbb Z}f\left(x+k\right)$$ is uniformly convergent.
This is what I attempted: let
$$s_n\left(x\right)=\sum_{k=-n}^nf\left(x+k\right).$$
Then
$$\left|s_m\left(x\right)-s_n\left(x\right)\right|=\left|\sum_{n<\left|k\right|\leq m}f\left(x+k\right)\right|\leq\sum_{n<\left|k\right|\leq m}\left|f\left(x+k\right)\right|.$$
However, I do not know how to make the right hand side arbitrarily small to show that $\left(s_n\left(x\right)\right)_{n\in\mathbb N}$ is uniformly Cauchy. Moreover, I do not know how to use the fact that $x^2f\left(x\right)$ is bounded.
Let's write $f_k=f(x+k)$. The series $\sum_{k\in \mathbb{Z}} f_k$ cannot converge uniformly on $\mathbb{R}$ unless $f$ is identically zero. Indeed, for uniform convergence of $\sum f_k$ it is necessary (though not sufficient) that $\sup |f_k|\to0$ as $|k|\to\infty$. Clearly, $\sup_{\mathbb{R}} |f_k| = \sup_{\mathbb{R}}|f|$ is independent of $k$.
But it's true that the series converges uniformly on every bounded interval, say $[-A, A]$. Continuity and integrability are not needed for this: it suffices to know that there is $M$ such that $x^2|f(x)|\le M$ for all $x$. Indeed, by the triangle inequality $|x+k|\ge |k|-A$ for $x\in [-A,A]$, which implies $$ \sup_{[-A,A]} |f_k| \le \frac{M}{(|k|-A)^2},\quad \text{for }|k|>A, $$ hence the series converges uniformly by the Weierstrass M-test.