As a function, $T: \mathbb{R}^{2\times2}\longrightarrow\mathbb{R}^{2\times2}$. You can check that $T$ is linear, so that makes it a linear transformation.
In general, consider the following. Let $U,V$ be finite dimensional vector spaces and $F:U\longrightarrow V$ be a linear map. Given an ordered basis $\alpha$ of $U$ and an ordered basis $\beta$ of $V$, we may write down a matrix $M=M_F(\alpha,\beta)$ that represents the map $F$ in the following sense:
Each $u\in U$ can be written as a linear combination of the elements of $\alpha$ in a unique way; the coefficients of this linear combination are the coordinates of $u$ in base $\alpha$, and we call these $[u]_\alpha$. Analagously, we may do the same for $v \in V$ with $\beta$, and call these coordinates $[v]_\beta$.
Example: Consider the canonical basis of $\mathbb{R}^2$ and let $u=(2,6)$ in this basis. Now, let $\alpha=\{(1,1),(1-1)\}$ be another basis for $\mathbb{R}^2$. Notice the basis is ordered, with $\alpha_1=(1,1)$ and $\alpha_2=(1,-1)$. We have that $u=4\alpha_1-2\alpha_2$, so $[u]_\alpha={(4,-2)}$.
The matrix $M$ is such that $M\cdot{[u]}_\alpha={[Fu]}_\beta$, meaning that when you multiply a vector $u$ written in base $\alpha$ by $M$, you get $Fu$ written in base $\beta$. We will construct $M$ one column at a time.
Indeed, let $e_i=(0,\dots,0,1,0,\dots,0)$, where the $1$ appears at position $i$. You can check that for any matrix $A$, $Ae_i$ is simply the $i$-th column of $A$. Now, how does $M$ act on $e_i$? Observe that $e_i={[\alpha_i]}_\alpha$, hence
$$Me_i=M{[\alpha_i]}_\alpha=[F\alpha_i]_\beta$$
In words, this means that the $i$-th column of $M$ is obtained by taking the $i$-th element of base $\alpha$, applying $F$ to it, then writing it down in base $\beta$.
Now, with all of this out of the way, let's tackle the problem at hand.
Since $\mathbb{R}^{2\times2}$ is four-dimensional, the matrix $T(B_v,B_w)$ will be $4\times4$. We name the matrices of $B_v=B_w\,$,
\begin{align}
&b_1=\left[\begin{array}{l}1&0\\0&0\end{array}\right]&&b_2=\left[\begin{array}{l}0&1\\0&0\end{array}\right]\\
&b_3=\left[\begin{array}{l}0&0\\1&0\end{array}\right]&&b_4=\left[\begin{array}{l}0&0\\0&1\end{array}\right]
\end{align}
so that $B_v=B_w=\{b_1,b_2,b_3,b_4\}$, in this order. Now, remember, the $i$-th column of $T(B_v,B_w)$ will be the $i$-th element of $B_v$, transformed by $T$, and then written in base $B_w$, that is
The $i$-th column of $T(B_v,B_w)$ is $Tb_i$ written in base $B_w$
Calculating the $Tb_i$'s:
\begin{align}
&Tb_1=\left[\begin{array}{l}1&0\\0&0\end{array}\right]&&Tb_2=\left[\begin{array}{l}0&\frac12\\\frac12&0\end{array}\right]\\
&Tb_3=\left[\begin{array}{l}0&\frac12\\\frac12&0\end{array}\right]&&Tb_4=\left[\begin{array}{l}0&0\\0&1\end{array}\right]
\end{align}
Now, writing them in base $B_w$:
\begin{align}
&[Tb_1]_{B_w}=(1,0,0,0)&&[Tb_2]_{B_w}=\left(0,\frac12,\frac12,0\right)\\
&[Tb_3]_{B_w}=\left(0,\frac12,\frac12,0\right)&&[Tb_4]_{B_w}=(0,0,0,1)
\end{align}
Finally:
$$T(B_v,B_w)=\pmatrix{1&0&0&0\\0&\frac12&\frac12&0\\0&\frac12&\frac12&0\\0&0&0&1}$$
