If $U\subset W$ then every function in $L^p (U)$ can be extended to a function in $L^p (W)$, for example by setting it to be 0 outside of $U$.
However, not every continuous or differentiable function on $U$ can be extended to a continuous or differentiable function on $W$. For example, $1/x$ on $(0,1)$ cannot be extended continuously and $\sqrt{x}$ on $(0,1)$ cannot be extended to a function that's differentiable on $(-1,1)$.
I am learning about Sobolev spaces and am wondering whether it is true that any $f\in W^{1,1}(U)$ can be extended to a function in $W^{1,1}(W)$?
This depends on what $U$ is; the term Sobolev extension domain was introduced for such $U$. To see why this matters, take $U$ to be the unit disk in $\mathbb{R}^2$ with a radial slit. Then a function that has different boundary limits on two sides of the slit cannot be extended to a $W^{1,1}$ function on $\mathbb{R}^2$, due to the lack of ACL property across the slit. Assuming that $U$ is a Jordan domain doesn't solve the problem, because one can create essentially the same problem with an inward cusp: an extension would need to have large gradient across the cusp, putting it out of $W^{1,1}$.
Nice domains do have the Sobolev extension property: Smooth, Lipschitz, uniform, locally uniform... For an overview of the topic, I recommend the Ph.D. thesis of Luke Rogers, in particular Chapter 1 which is introductory.