Can a matrix transformation ever make a linearly dependent matrix linearly independent? For example, if...

Step 1
No it can not:
${\displaystyle A\in K^{m\times n},B\in K^{p\times m}}$
A is linear dependent, so there exist ${\displaystyle {\lambda }_k\in K}$ not all equal to 0 with
${\displaystyle \sum _{k=1}^n{\lambda }_k{a}_k=0.}$
where ${\displaystyle A=(a_1,a_2,\dots ,a_n)}$ or: there is a non-zero vector ${\displaystyle x_0=\left({\lambda }_k\right)\ne 0}$ with
${\displaystyle A{x}_0=0.}$
For BA we have
${\displaystyle \left(BA\right)x_0=B\left(A{x}_0\right)=B0=0}$
so regardless of the choice of B the vector ${\displaystyle x_0}$ is a non-zero kernel vector for BA, so BA can not be linear independent.

Step 1
Let the matrices A and B be given. Suppose the columns of A are the column vectors ${\displaystyle A_i}$ where i runs over the number of columns, let that number be k.
Now,
${\displaystyle BA=[B{A}_1\left|B{A}_2\right|\dots B{A}_k]}$
So, your question boils down to asking can B act on the set of linearly independent vectors ${\displaystyle \{A_1,A_2,\dots A_k\}}$ such that the resulting set ${\displaystyle \{B{A}_1,B{A}_2,\dots B{A}_k\}}$ is linearly dependent.
The answer is yes, let B be the transformation that mixes the columns of A, for example, define ${\displaystyle B{A}_1=A_1+A_2}$ and ${\displaystyle B{A}_i=A_i}$ for ${\displaystyle i\ge 2}$