Basis vectors and non-square matrix transformations

Suppose we have a $$2\times3$$ matrix which takes a $$3\times1$$ column vector expressed in a basis $$\mathcal{B}$$. The output of this multiplication is a $$2\times1$$ vector. That is: $$\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}_\mathcal{B} = \begin{bmatrix} x_1' \\ x_2' \end{bmatrix}_\text{basis?}$$ My question is what is the basis of this new vector?

If we take the vector: $$\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}_\mathcal{B}$$ then it sends it in: $$\mathbf{x}= \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$$ Of course we can write $$\mathbf{x}$$ as: $$\mathbf{x} = a_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + a_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

but we don't know the basis.

You are basically asking given a finite spanning set $$\mathcal{A}$$, how one can find a subset of it that is a basis, one way to do this would be to start with the empty set and keep adding elements until the set you get spans the space(and then you stop).
Let $$\{ v_1, v_2, v_3 \}$$ be the basis $$\mathcal{B}$$, and $$T$$ be the linear transformation represented by the multiplication. A preimage of $$T$$ is some vector $$x_1 v_1 + x_2 v_2 + x_3 v_3$$, where $$x_1, x_2, x_3 \in \mathbb{R}$$ (or whatever field is relevant). The image is expressed with respect to a basis $$\{w_1, w_2\}$$, let's call it $$\mathcal{C}$$. This basis is such that $$T v_1 = a_1 w_1 + a_2 w_2$$, $$T v_2 = b_1 w_1 + b_2 w_2$$, and $$T v_3 = c_1 w_1 + c_2 w_2$$.