I assume everything is considered as a real vector space. The span is characterized as the set of linear combinations of elements. So the span of these matricies, call them $A=T(1),B=T(x)$, are the sums $rA+sB$ where $r,s \in \mathbb{R}$. This is equivalent to the set $$\bigg\{M\in \mathcal{M}_{2\times 2}: M=\begin{pmatrix} -s & 0 \\ 0 & r\\ \end{pmatrix}; r,s\in \mathbb{R}\bigg\}$$
Since every $r,s$ is possible, this is the set of all $2\times 2$ diagonal matrices.
If we note that $A=T(1)$, $B=T(x)$, $3B=T(x^2)$ we can further write for any polynomial given by $P(x)=a+bx+cx^2$ that $T(P(x))=aA+bB+3cB$ and write explicitly what this matrix is in the image.
Without having worked through the course myself, this appears to be more a product of confusing notation than of a deeply rooted misunderstanding.
Instead, considering the following system:
$$
1a + 2b = 2 \\
−3a + 4b = 5
$$
The goal of creating vectors here is that we want to be able to write the set of equations above as $aA + bB = C$ for some column vectors $A, B, C$. If you think about it in those terms, then it becomes more straightforward why you need to define:
$$
A=\begin{bmatrix}1 \\ -3\end{bmatrix}
B=\begin{bmatrix}2 \\ 4\end{bmatrix}
C=\begin{bmatrix}2 \\ 5\end{bmatrix}
$$
Now we can re-write this system as: $aA + bB = C$.
$$
\begin{bmatrix}1a \\ -3a\end{bmatrix} +
\begin{bmatrix}2b \\ 4b\end{bmatrix} =
\begin{bmatrix}2 \\ 5\end{bmatrix}
$$
It is not so much that these vectors represent vectors in the traditional plane, but rather as a concise way to represent the system of equations. Ultimately, people will begin to express these not as equations in vectors, but as augmented matrices, which nicely summarize the system of equations.
$$
\left[\begin{array}{cc|c}
1 & 2 & 2 \\
-3 & 4 & 5
\end{array}\right]
$$
It is my opinion that suspending parallels to the $\mathbb{R}^2$ plane as soon as possible when studying linear algebra tends to be a worthwhile endeavour. It can be useful when drawing parallels, but ultimately hampers the ability to think more broadly about the concepts.
Best Answer
Notice that $U$ and $W$ are linear independent, so $span \{U,W\}=\mathbb{R}^2$and $\begin{bmatrix} H\\ K \end{bmatrix} \in \mathbb{R}^2$ for all $H,K \in\mathbb{R}$.
$W$ and $U$ are linear independent, because $W \neq \alpha U$ and $U \neq \alpha W$ for all $\alpha \in\mathbb{R}$.