The linear combinations of the vectors $v,w$ can be denoted $span(v,w)=kv+cw$, where $v,w \in V$ and $k,c \in \mathbb{F}$ (scalars).
If the two vectors are linearly independent, informally consider the follow scenario:
Let $V = \mathbb{R^2}$. Let two vectors $v,w \in V$. Set $v = (x,0)$ and $w = (0,y)$. There is no scalar $k \in \mathbb{F}$, that when multiplied by the vector $v = (x,0)$, yields a vector of the form $w = (0,y)$. The two independent vectors, when combined linearly as $kv+cv$, make up a whole plane or $\mathbb{R^2}$. You can think of these vectors as the $x-axis$ and $y-axis$ coming together to form the $xy-plane$. When taking the set of all linear combinations of the vectors $v$ and $w$ we can obtain any other vector of the form $(x,y)$, and thus $span(v,w) = \mathbb{R^2}$.
Formally, the only combination $kv+cw$ that yields the $0$ vector is by setting each $k,v \in \mathbb{F}$ equal to $0$. Is is clear that the only scalars that satisfy the linear combination $k(x,0) + c(0,y) = (0,0)$ for the $0$ vector are $k = c = 0$, which implies that the two vectors in $\mathbb{R^2}$ are independent and span the plane.
Now, consider a separate case, where the vectors $v,w \in V$ are $(x,0)$ and $(2y,0)$, respectively. Let $span(v,w) = a(x,0) + b(2y,0)$ which denotes all possible linear combinations of the vectors. However, it is apparent that the $0$ can be obtained by more than just the "obvious" way which means setting $a = b = 0$ to obtain $(0,0) + (0,0) = (0,0)$. Setting $y = \frac{-1}{2}x$, and we get $a(x,0)+b(-x,0)$ which equals $0$ for $a = b = 0$ or $a = b = 1$, which shows that the two vectors are dependent and as a result do not span $\mathbb{R^2}$, so their linear combinations do not fill a plane.
It's not that $v$ and $w$ can't lie on the same line. It's just that if we have two vectors are on the same line, then they are two linearly dependent vectors in $\mathbb{R^2}$, and as a result, do not fill a plane.
The subspace $V$ has dimension $3$ inside $\mathbb{R}^4$ (thus it is a hyperplane). Since $V \subseteq V+U \subseteq \mathbb{R}^4$, the first subspace has dimension $3$ and the last subspace has dimension $4$, we must have either that $\operatorname{dim} V+U = 3$ -- and thus $V+U = V$ -- or $\operatorname{dim} V+U = 4$ -- and thus $V+U = \mathbb{R}^4$. Moreover, $V+U = V$ holds if and only if $U \subset V$.
So it all boils down to whether $U$ is contained in $V$. This is immediate to check: does every vector in $U$ have last coordinate equal to $0$? Clearly not -- look at $u_2$ -- so $U+V = \mathbb{R}^4$.
Added: With regard to the "supplementary": 1) Previously I had a "$W"$ in my answer. That was a typo: all instances of it have been changed to $U$. 2) Yes, if $U+V = \mathbb{R}^4$ then $\{e_1,e_2,e_3,e_4\}$ is a basis for $U+V$. You asked how one would know that $U+V = \mathbb{R}^4$, so my answer gave a more detailed and motivated argument for that. Your work doesn't make much sense to me, since you start by writing that the set $U+V$ has just two elements $s_1$ and $s_2$. Clearly this is not the case. Nor is it clear what $s_1$ and $s_2$ mean, since you have unspecified variables $x_1$, $x_2$, $x_3$ in them.
Best Answer
Generally when people talk about planes in three dimensions, they mean a set of the form $P=\{ x \in \mathbb{R}^3 | n^T x = d\}$, where $n \in \mathbb{R}^3$ is a fixed non-zero vector and $d \in \mathbb{R}$.
A plane is infinite in extent as $\ker n^T$ is two dimensional, and the point $x_0={d \over \|n\|^2} n \in P$, so $x_0+k \in P$ for any $k \in \ker n^T$.
If the plane passes through the origin, it is easy to see that the corresponding $d$ above must be zero.
The intersection of two planes that pass through the origin will be given by $L = P_1 \cap P_2$, where $P_k = \{ x \in \mathbb{R}^3 | n_k^T x = 0\}$. We can write $L = \{ x \in \mathbb{R}^3 | n_1^T x = 0, n_2^T x = 0\} = \ker N^T$, where $N = \begin{bmatrix} n_1 & n_2 \end{bmatrix}$.
If the planes are different (equivalently $n_1, n_2$ do not lie on the same line), then $n_1, n_2$ are linearly independent, and $\dim \ker N^T = 1$, hence $L$ is a line of infinite extent passing through the origin.
You can always choose subsets $S_1, S_2$ of $P_1, P_2$ so that $S_1$ and $S_2$ only intersect at the origin, but then $S_1,S_2$ will not be planes in the usual sense.