Trouble understanding Sum of Subspaces
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 $(19,0,0)+(0,-3,0)= (19,-3, 0)$.
First, you can show that the sum of subspaces is also a subspace directly from definition. For example if $$ u_{1}+...+u_{m}\in U_{1}+...+U_{m} $$
where $u_{i}\in U_{i}$ and $$ u_{1}'+...+u_{m}'\in U_{1}+...+U_{m} $$
where $u_{i}'\in U_{i}$ then $$ u_{1}+...+u_{m}+u_{1}'+...+u_{m}'=(u_{1}+u_{1}')+...+(u_{n}+u_{n}') $$
and since $U_{i}$ is a subspace it is closed under addition and so $u_{i}+u_{i}'\in U_{i}$ and thus the above sum is in $U_{1}+...+U_{m}$.
Regarding your second example: $U+W=\{(x,y,0)|\, x,y\in\mathbb{R}\}$ is as straightforward as it seems, the part that looks a bit strange is to find $U+Z$ and lets work by definition to understand what it is:
$$ U=\{(x,0,0)|\, x\in\mathbb{R}\} $$ $$ Z=\{(y,y,0)|\, y\in\mathbb{R}\} $$
If $$ u+z=(a,b,c) $$
then clearly $c=0$. We claim that if $(x_{0},y_{0})\in\mathbb{R}^{2}$ then there are some $u\in U,z\in Z$ s.t $a=x_{0},b=y_{0}$- Indeed $u+z$ is of the form $$ (x,0,0)+(y,y,0)=(x+y,y,0)=(x_{0},y_{0},0) $$
then we have the solution $$ y=y_{0} $$ $$ x=x_{0}-y_{0} $$
and if you wish to verify $$ (x_{0}-y_{0},0,0)+(y_{0},y_{0},0)=(x_{0},y_{0},0) $$
So $$ \{(x,y,0)|x,y\in\mathbb{R}\}\subseteq U+Z $$
and the other containment is clear and so the two sets are in fact equal.
I hope this makes things clear, please comment if not
The sum of $U_1$ and $U_2$ contains anything that can be obtained by adding something from $U_1$ to something from $U_2$.
You say you understand why $U+W$ in this example is equal to $\{(x,y,0) : x,y\in \Bbb R\}$, so it seems that you're puzzled about why $U+Z$ is also equal to this.
The sum of $U$ and $Z$ is anything that can be obtained by adding something from $U$ to something from $Z$. The claim is that this is equal to $U+W$. To be equal, the two spaces must contain the same vectors. So the claim is that we can get any vector $(x,y,0)$ from $U+W$ by adding something from $U$ to something from $Z$, and also that we can only get vectors of that form.
So you need to show:
- If you add something from $U$ to something from $Z$, the sum must have the form $(x,y,0)$, which will show that it is in $U+W$.
- Any vector $(x,y,0)$ that is in $U+W$ can also be obtained by adding something from $U$ to something from $Z$.