The matrix corresponding to $\;T\;$ and the given basis is in fact
$$\begin{pmatrix}2&-2&-2&-2\\0&0&0&0\\0&0&0&0\\-1&1&1&1\end{pmatrix}$$
From this we can see the matrix rank = the transformation's image's dimension, is one, and thus its kernel (both of the matrix and the transformation) has dimension three.
If you don't understand this I can't see a way to explain you any further without solving completely the question and you don't understanding a thing...
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be a linear map. Suppose that $A\in \mathbb{R}^{m\times n}$ is the matrix of $f$ w.r.t. the standard bases. Since the rank of $A$ is equal to the column rank of $A$, it suffices to show that the columns of $A$ vectors in the image of $f$. Now let $e_i$ be the $i$-th standard basis vector of $\mathbb{R}^n$. Then $f(e_i)=Ae_i=A_i$ where $A_i$ is the $i$-th column of $A$. This completes the proof.
Can you figure what happens the matrix of $f$ is not given w.r.t. the standard bases.?
Edit: You edited the question. The matrix $A$ is no longer the matrix of $f$ w.r.t. the standard bases. The above argument fails since the columns of $A$ do not necessarily belong to the image. However, the columns represent the coordinates of images of basis vectors. So what does belong to your image?
Second Edit: Let's work with your example. You know that basis of $f$ w.r.t. the bases $B$ en $B'$. What information does give us? We know that $f((1,3))=2(0,0,1)+1(1,0,-1)-1(0,1,0)$. Notice that the coordinates appearing in this last expressions are $(2,1,-1)$ which is exactly the first column of $A$. If we want to write $(1,3)$ w.r.t. the basis $B$, then we bet $(1,3)=1(1,3)+0(1,2)$. Notice that $A\begin{pmatrix}1\\0
\end{pmatrix}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$. This result is the coordinate of $f((1,3))$ w.r.t. the basis $B'$. We do not have that $(2,1,-1)\in \text{Im}(f)$, but $f((1,3))=2(0,0,1)+1(1,0,-1)-1(0,1,0)=(1,-1,1)\in \text{Im}(f)$. In the same fashion $f((1,2))=-1(0,0,1)+3(1,0,-1)+1(0,1,0)=(3,1,-4)\in \text{Im}(f)$. Hence $\text{Im}(f)=\text{Span}\left\{(1,-1,1),(3,1,-4)\right\}$.
Best Answer
let me write $e_1 = (1,0)^\top, e_2 = (0,1)^\top.$ then you have $$Ae_1 = 0, Ae_2 = (1,2,1)^\top.$$ this means $e_1$ is in the $\ker(A)$ and $(1, 2,1 )^\top$ is in the $image(A).$ we also have sum of the dimensions of $\ker(A) $ and $image(A)$ is $2.$ therefore $$\ker(A) = span\{e_1\}, image(A) = span\{(1, 2, 1)^\top\}. $$