Let's suppose $W=U_1\oplus U_2$ in the sense you have defined in your first paragraph.
Note that
$$
\underbrace{0}_{\in W}=\underbrace{0}_{\in U_1}+\underbrace{0}_{\in U_2}
$$
so by the uniqueness of the representation on the RHS above, if you have
$$
\underbrace{0}_{\in W}=\underbrace{u_1}_{\in U_1}+\underbrace{u_2}_{\in U_2}
$$
then $u_1=0$ and $u_2=0$.
The answer to your first question is no. Take the real vector space $\mathbb{R}^2$.
Then $\mathbb{R}^2 = \operatorname{span}\{(1,0)\}\oplus \operatorname{span}\{(0,1)\} = \operatorname{span}\{(1,0)\}\oplus \operatorname{span}\{(1,1)\}$
but $\operatorname{span}\{(0,1)\}\neq \operatorname{span}\{(1,1)\}.$
It is also not true that the direct sum of two spaces is equal to the entire space. Indeed, take any vector space of dimension $d \geq 1$. Then
$$\{0\}= \{0\} \oplus \{0\}$$
is not equal to the entire space.
A less trivial example:
$$\operatorname{span}\{(1,0,0)\}\oplus \operatorname{span}\{(0,1,0)\} = \mathbb{R} \times \mathbb{R} \times \{0\} \neq \mathbb{R}^3$$
Best Answer
The direct sum depends on context. There are two notions of direct sum: the inner direct sum and the outer direct sum.
With the inner direct sum, we have some large vector space and two subspaces. We must assume that the intersection of the two subspaces is zero in order to form the inner direct sum.
With the outer direct sum, we take two vector spaces that have nothing to do with each other and slam them together. I.e. you get a vector in the outer direct sum by appending vectors from your two vector spaces.
It turns out that these two notions are isomorphic. That is, if you take the inner direct sum of two subspaces in a big vector space, that is isomorphic to the vector space obtained by taking the outer direct sum of those subspaces as vector spaces in their own right.
In your example, we are viewing the two copies of $\mathbb{R}$ as distinct vector spaces that have nothing to do with each other and then taking the outer direct sum of them.
To think about it in terms of the inner direct sum, think of one of the copies of $\mathbb{R}$ as the x-axis and the other copy of $\mathbb{R}$ as the y-axis. Then their intersection is zero, and taking the inner direct sum yields $\mathbb{R}^2$.