Here it is very easy to see how the standard basis vectors can be written as linear combinations of the vectors in B.
Example:
$(1,0,0)=(1,1,0)-(0,1,0)$
Therefore $T((1,0,0))=T((1,1,0))-T((0,1,0))$
Moreover we know that $T((1,1,0))=2(1,1,0)+n(0,1,0)=(2,2+n,0)$ and that $T((0,1,0))=(m,m,0)$, whence $T((1,0,0))=(2-m,2+n-m,0)$.
This is the first column in your sought after matrix, and you can do similarly with the other two standard basis vectors.
The more general approach is to first find the change of basis matrices $P_{\mathcal{B}\rightarrow{\mathcal{E}}}$ and $P_{\mathcal{E}\rightarrow{\mathcal{B}}}$, where $\mathcal{E}$ denotes the standard basis. One then has the following formula:
$[T]_{\mathcal{E}}=P_{\mathcal{E}\rightarrow{\mathcal{B}}}[T]_{\mathcal{B}}P_{\mathcal{B}\rightarrow{\mathcal{E}}}$
If you think of the left null space of $M$ as consisting of row vectors, and you also think of the elements of the annihilator of a subspace as being row vectors, then the left null space of $M$ is the annihilator of the range of $M$.
Here is a proof. Let $z$ be a column vector in $\mathbb R^m$. Then
\begin{align}
z^T \in R(M)^\circ &\iff z^T Mx = 0 \quad \text{for all } x \in \mathbb R^n\\
&\iff (M^T z)^T x = 0 \quad \text{for all } x \in \mathbb R^n\\
&\iff M^T z = 0 \\
&\iff z^T \text{ is in the left null space of $M$.}
\end{align}
Here is a generalization of the above fact.
Let $V$ and $W$ be finite dimensional vector spaces over a field $F$, and let $T:V \to W$ be a linear transformation. Let $V^*$ and $W^*$ be the dual spaces of $V$ and $W$ (respectively) and let $T^*:W^* \to V^*$ be the dual of $T$, defined by
$$
\langle T^*z, x \rangle = \langle z, Tx \rangle.
$$
Then the annihilator of the range of $T$ is the null space of $T^*$:
$$
R(T)^\circ = N(T^*).
$$
This is a generalization of the "four subspaces theorem" emphasized in Gilbert Strang's linear algebra books. (This theorem is sometimes called the "fundamental theorem of linear algebra".)
Here's a proof:
\begin{align}
z \in R(T)^\circ &\iff \langle z, Tx \rangle = 0 \quad \text{for all } x \in V\\
&\iff \langle T^* z, x \rangle = 0 \quad \text{for all } x \in V \\
&\iff T^*z = 0 \\
&\iff z \in N(T^*).
\end{align}
Best Answer
No. The annihilator of $U$ is a subset of $(\Bbb R^3)^*$, not of $\Bbb R^3$. You have$$U^\circ=\{\alpha\in(\Bbb R^3)^*\mid\alpha(1,0,0)=0\},$$a basis of which is $\{\alpha_1,\alpha_2\}$, with$$\alpha_1(x,y,z)=y\text{ and }\alpha_2(x,y,z)=z.$$And, yes,$$\alpha_1(x,y,z)=(x,y,z).(0,1,0)\text{ and }\alpha_2(x,y,z)=(x,y,z).(0,0,1).$$Nevertheless, $\alpha_1,\alpha_2\in(\Bbb R^3)^*$, whereas $(0,1,0),(0,0,1)\in\Bbb R^3$.