A point that might confuse you is that the same letters are used, let us rewrite this.
If you have some element from $U+W$ then it is of the form $(x,y,0)$ for some $x,y$.
If you have some element from $U+Z$ then it is of the form $(v+w,w,0)$ for some $v,w$.
Now the claim is that these two really yield the same collection of elements. I assume you can see that each element of the latter form is of the former. More specifically if we pick a generic element $ (v+w, w,0) $ in $U+Z$, we can equivalently express the element in the form $ (x,y,0) $ by setting $ x = v + w$ and $ y = w $. This shows that $ U + Z \subset U + W$.
For the other direction we can select an arbitrary element $(x, y, 0) $ of $ U + W $ and re-express it in the form $(v+w,w,0)$ by setting $ v = x - y $ and $ w = y $. The latter shows $ U + W \subset U + Z$.
The two subset relations $U+Z \subset U+W $ and $ U + W \subset U + Z$ imply that $U+Z = U+W$.
Added following the edit: a single element of the sum $U+W$ is one element of $U$ plus one element of $W$. And $U+W$ is the set of all these elements together.
So, you have since $(19,0,0)$ in $U$ and $(0,-3,0)$ in $W$ that $(19,0,0)+(0,-3,0)$ in $U+W$. However, you can of course evaluate that sum $(19,0,0)+(0,-3,0)= (19,-3, 0)$.
[…] and second part, $U_1, \dotsc, U_m$ must contain $U_1 + \dotsb + U_m$.
This is wrong. Consider for example $V = \mathbb{R}^2$ and the subspaces
$$
U_1 = \{(x,0) \mid x \in \mathbb{R}\}
\quad\text{and}\quad
U_2 = \{(0,y) \mid y \in \mathbb{R}\}.
$$
Then $U_1 + U_2 = \mathbb{R}^2$, but neither $U_1$ nor $U_2$ contain $\mathbb{R}^2$.
The statement that we really want is the following:
$U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$, and if $W \subseteq V$ is any subspace with $U_1, \dotsc, U_n \subseteq W$, then we already have $U_1 + \dotsb + U_n \subseteq W$.
You already understand the first part of this statement, that $U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$.
For the other part suppose $W \subseteq V$ is some subspace containing $U_1, \dotsc, U_n$. For all $u_1 \in U_1, \dotsc, u_n \in U_n$, we then have $u_1, \dotsc, u_n \in W$. Because $W$ is a subspace it follows that also $u_1 + \dotsb + u_n \in W$ (because $W$ is closed under finite sums). Therefore
$$
U_1 + \dotsb + U_n
= \{u_1 + \dotsb + u_n \mid u_1 \in U_1, \dotsc, u_n \in U_n\}
\subseteq W.
$$
Best Answer
If $V$ is a subspace containing $U_1,\ldots,U_m$, then $V\supset U_1+U_2+\cdots+U_m$. In particular, the smallest subspace containing $U_1,\ldots,U_m$ will also contain $U_1+U_2+\cdots+U_m$. But $U_1+U_2+\cdots+U_m$ is a subspace itself, which contains $U_1,\ldots,U_m$. So, $U_1+U_2+\cdots+U_m$ contains the smallest subspace containing $U_1,\ldots,U_m$, and therefore they're equal.