For finding an inverse matrix using elementary transformations, starting with $AA^{-1}=I$, we only use row transformations to get to the following equation $IA^{-1}= B$
And if we start with $A^{-1}A=I$ then we only use coloumn transformations to get to the following equation $A^{-1}I= B$. Why is this? Can't I use both row and column transformations in the same equation and just apply it to both the right hand side as well as the left hand side?
You can use a sequence of linear transformations $(T_{i})_{i=1}^{n}$, such that:
\begin{eqnarray*} A X = I \text{ applied }T_{1}&\Rightarrow& T_{1} A X = T_{1} I = T_{1}\\ &\Rightarrow& T_{2} T_{1} A X = T_{2}T_{1}\\ &\Rightarrow& T_{n} \cdots T_{1} A X = T_{n} \cdots T_{1}\\ \end{eqnarray*} if $T_{n}\cdots T_{1}$ is such that $T_{n}\cdots T_{1} A = I$, then $T_{n}\cdots T_{1} = A^{-1}$. Then
\begin{eqnarray*} &\Rightarrow& \underbrace{T_{n} \cdots T_{1} A}_{= I} X = T_{n} \cdots T_{1}\\ &\Rightarrow& X = T_{n} \cdots T_{1} = A^{-1} \end{eqnarray*}
Therefore, when applying operations on the matrix, you are applying linear transformations, this means that you multiply the matrix $T_ {i}$ in the i-th operation, then what you have left is the inverse of the matrix $A$ which turns out to be the sequence of matrices of the applied applications.
The idea can be applied to the right or to the left, and you will get row operations or column operations as a result.