The image vectors $\{T(v_{1}),\ldots, T(v_{n})\}$ can form a basis only if the kernel of $T$ is zero. This is not true anymore if $\{w_{1},\ldots,w_{n}\}$ are linearly dependent, because by definition you can find scalars $\lambda_{i}$'s not all zero such that
$$ \lambda_{1}w_{1}+\cdots+\lambda_{n}w_{n}=0 $$
Then $T(\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n})=0$, so we have a non-zero element in the kernel (if $\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n}=0$ then all $\lambda_{i}$'s would be zero by linear independence of the $v_{i}$'s).
So in the first theorem you need to assume that $T$ is an isomorphism (or equivalently that it has trivial kernel if both spaces have the same dimension).
You can't define a linear transformation from $\mathbb{R}^2$ to another space by defining only the image of one vector. You have to define it on a basis of the vector space.
For sure you can define a linear transformation $T$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ such that $T\bigl([1,1]\bigr)=[2,0,2]$. It is sufficient to complete $[1,1]$ to a basis of $\mathbb{R}^2$ with another vector $v$, linearly independent from $[1,1]$ (for instance with the vector $[1,0]$) and define the image of $v$. In whatever way you define the image of the second vector of the basis you obtain a linear function.
Of course the way you define $T$ on another basis' vector will change the function, but for sure it will be a linear function such that $T\bigl([1,1]\bigr)=[2,0,2]$.
What you did while trying to verifying the linearity of your function is forcing other properties to the function: why should $T([u_1+v_1, u_2+v_2])=[2(u_1+v_1), 0, 2(u_2+v_2)]$? By the only property you stated at the beginning you don't have this equality. A function which maps $[1,1]$ to $[2,0,2]$ is not forced to be linear. Think for instance to the constant function $T(v)=[2,0,2] \ \forall \ v \in \mathbb{R}^2$.
An example of a linear function $T$ which satisfies your property is the following:
Take the standard basis of $\mathbb{R}^2$, let's say $B=\bigl\{[1,0],[0,1]\bigr\}$. If you define $T$ by $T\bigl([1,0]\bigr)=[2,0,0]$ and $T\bigl([0,1]\bigr)=[0,0,2]$, then, by linearity of $T$ you have:
$$T\bigl([1,1]\bigr) \ = \ T\bigl([1,0]\bigr)+T\bigl([0,1]\bigr) \ = \ [2,0,0]+[0,0,2]=[2,0,2] \ .$$
This function is represented by the following matrix (the columns are the images of the vectors of the standard basis):
$$\left[\begin{matrix} 2 & 0 \\ 0 & 0 \\ 0 & 2 \end{matrix} \right] \ .$$
You can also swap the images of the two basis vectors and you obtain the same result but a different function.
Best Answer
You are right. You can remind your teacher that every matrix generates some linear transformation. In our case it is $$ A=\begin{pmatrix} 1 &0 \\ 1 &0 \end{pmatrix}. $$ Certainly $$ A\cdot\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} v_1 \\ v_1 \end{pmatrix}. $$ I am assuming that $(v_1,v_2)$ are the coordinates of a vector in two-dimensional space.