How to show that a mapping is linear 
Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T\begin{pmatrix}u_1\\u_2\end{pmatrix} = \begin{pmatrix}u^2_2\\2\\u_1-u_2\end{pmatrix}$
Show T is linear.

How can I do this using the rules I know
$i)  \, T(u +v) = T(u) + T(v)$
$ii) \,  T(ru) = rT(u)$ ?
For example, in this case what is $u$ and $v$ for me?
UPDATE

Let $S: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T(\begin{pmatrix}u_1\\u_2\end{pmatrix}) = \begin{pmatrix}u_2^2\\2\\u_1-u_2\end{pmatrix}$. Show $S$ is not linear

So I think I've got a solution but want to double check that this answer will suffice. (By the way I let $T = S$
I know that I can use the property
$T(ru) = rT(u)$ and let $u = 0$ to get $T(0) = \begin{bmatrix}0\\2\\0\end{bmatrix} \ne 0 \implies$ not linear. Isn't that all I really need to prove that $S$ is not linear
 A: To answer your question:
$u,v$ are just elements of a vector space, in your case it is $\mathbb R^2$, that means
$$
u=(u_1,u_2),v=(v_1,v_2) \text{ with } u_1,u_2,v_1,v_2\in \mathbb R
$$
$r$ on the other hand is called a scalar and in your case the corresponding filed (over which your vector space is defined) is again $\mathbb R$, so $r\in \mathbb R$.
The following was an answer to the question before editing:
Now just check for example what @Bye_World already mentioned in the comments to see that 
$$
T(ru) = rT(u)
$$
fails (or check whether the $0$ is mapped to $0$) or just check your first property with 
$$
u=(1,0),v=(0,1)
$$
which gives
$$
T(u+v)=T((1,1))=(1,2,0)\neq(0,2,1)+(1,2,-1)=(1,4,0)=T(u)+T(v)
$$
So $T$ is not a linear mapping.
The following is an answer to the question after editing
Here $T$ is indeed linear, the second property is obvious and for the first one we just check
\begin{align}
T(u+v)=T((u_1+v_1,u_2+v_2))&=(u_2+v_2,0,u_1+v_1-u_2-v_2)\\&=(u_2,0,u_1-u_2)+(v_2,0,v_1-v_2)=T(u)+T(v)
\end{align}
A: First, the "rules" you present are not rules, they are the definition of what is meant by "linear."  So they are the correct starting point.
Now you have to decide what addition and scalar multiplication mean when talking about members of $\Bbb{R}^2$ and $\Bbb{R}^3$ to check whether the two definiing properties will hold.  This, I think, is where your confusion comes in.  The problem obviously means to use the usual definitions of addition and scalar multiplication:
$$
\pmatrix{a\\b}+\pmatrix{c\\d}=\pmatrix{a+c\\b+d}\\
\pmatrix{a\\b\\c}+\pmatrix{d\\e\\f}=\pmatrix{a+d\\b+e\\c+f}\\
r\pmatrix{a\\b}=\pmatrix{ra\\rb}\\
r\pmatrix{a\\b\\c}=\pmatrix{ra\\rb\\rc}
$$
With that in mind, if $T$ is liniear then (taking $r = -1$ in the second property)
$$
\forall u_1,u_2: (-1)T\pmatrix{u1\\u2}=T\pmatrix{-u1\\-u2}
$$
But the left side is
$$
\pmatrix{-u_2^2\\-2\\u_2-u_1}
$$
and the right side is 
$$
\pmatrix{u_2^2\\2\\u_2-u_1}
$$
These are never equal, so $T$ is not linear.
A: You can also see this the way Xianjin Yang showed in the comments. We know $T(ru)=rT(u)$ for any $r \in \mathbb{R}$ and $u \in \mathbb{R}^2$. Let $r=0$ and we have $T(0)=0$ as a special case. But in this example, 
$$T(0)=\begin{pmatrix}0\\2\\0\end{pmatrix} \neq 0$$
Therefore, $T$ is not linear.
A: In this case, your $u$ and $v$ will be vector in the form $\begin{pmatrix} a\\b\end{pmatrix}$.
To check the first property, let $u = \begin{pmatrix} a\\b \end{pmatrix},v=\begin{pmatrix} c\\d \end{pmatrix}.$ Then we have
$$T(u+v)=T(\begin{pmatrix} a\\b \end{pmatrix}+\begin{pmatrix} c\\d \end{pmatrix})=T(\begin{pmatrix} a+c \\b+d \end{pmatrix})  \text{......(why?)}$$
You then compute $T(u) + T(v)$. If these two are equal then you are done for the first property. However, I am not saying that $T$ is a linear transformation since I haven't done the computation. But in fact you can tell that $T$ is not a linear transformation. This is a consequence of the following fact:
https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector
