Let $T: V \longrightarrow V$ be a linear transformation. If W is T invariant, and for another subspace of U of V, $V = U\bigoplus W$ is U L invariant?

Here $V$ is finite dimensional. I do not think that $U$ is necessarily $L$ invariant. We have each $v \in V$ is equal to $u + w$ for some $u \in U$ and $w \in W$. So $T(v) = T(u)+T(w) = u_1 + w_1$ for a $u_1$ and $w_1$ in $U$ and $W$ respectively. $T(w)$ is equal to something in $w$ but $T(u)$ is not necessarily.

However, if $T$ is one-to-one and onto. $T(W)$ maps to $W$ meaning $T(U)$ must map to $U.$ So it is invariant if the map is one-to-one and onto, but necessarily if it is not?

No, not even if the map is one-to-one and onto. Take $\mathbb{R}^2 = V$ and consider the operator $T$ given by the matrix

\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

Then $T$ is one to one and onto and it maps $(1,0)$ to itself. Therefore it maps the $x$ axis to itself which can serve as your $W$. There are actually many subspaces $U$ such that $W \oplus U = \mathbb{R}^2$ but not a single one is $T$ invariant.