For (1), the answer is almost always no; the combinatorial properties of open covers of $\mathbb{R}$ are far too restricted. Suppose that $M$ is compact and connected and such an $f$ and $\{U_\alpha\}$ exist. Let the image of $f$ be $[a,b]$. Each $U_\alpha$ is a union of disjoint intervals; by connectedness, only one of those disjoint intervals can intersect $[a,b]$. Restricting the codomain of $f$ to $[a,b]$, we may thus assume each $U_\alpha$ is a single interval. Now take a minimal subcollection of $\{U_\alpha\}$ that still covers $[a,b]$: this will be a finite sequence of intervals $I_k=(x_k,y_k)$ going from left to right such that only consecutive intervals intersect. That is, we have $$x_1\leq x_2\leq y_1\leq x_3\leq y_2\leq x_4\leq y_3 \leq\dots$$

Let $U$ be the union of the $f^{-1}(I_k)$ for $k$ odd and $V$ be the union of the $f^{-1}(I_k)$ for $k$ even. Then both $U$ and $V$ are disjoint unions of copies of $\mathbb{R}^n$, and they cover $M$. This implies that the LS-category of $M$ must be at most 1. In particular, for instance, this implies that the cohomology of $M$ can have no nontrivial cup products. If $M$ is not compact, a similar (but a little bit more difficult) argument can be made to reach the same conclusion.