To find the Col(A) we row reduce the matrix and get $R$ where $R$ is row equivalent to $A$. Now we choose the columns in $A$ corresponding to the pivot columns in $R$ as the basis. My question is when can we choose the same pivot columns in $R$ as a basis for $Col(A)$. What I think is
(A) If the matrix is Invertible, then we can choose from $A$ or $R$.
(B) If we have $m\times n$ matrix $A$ where $m<n$. and if $Rank(A)=m$, then we can choose either from $A$ or $R$.
Is my reasoning correct?
Your assumption (A) is correct. If $A$ is invertible then Col$(A)$ spans the whole space - as does Col$(R)$.
Your assumption (B) is also correct. The columns of $A$ are vectors in $K^m$ for the field $K$ in which we are working. That means that any $m$ linear independent columns of $A$ (which exist when rank$(A)=m$ will form a basis of the whole space $K^m$). For $R$ the same holds true because rank$(R)=$ rank$(A)=m$.
A little further inside: In both cases the possibility of choosing either from $R$ or $A$ comes from the fact that Col$(A)$ spans the whole vector space we are looking at - which will then be true for $R$ as well because the rank doesn't change. In general it is not possible to choose from $A$ or $R$ as you probably know. For example the matrix $$A= \begin{pmatrix} 1&2\\1&2 \end{pmatrix}$$ has column space $$\text{Col}(A) = \text{span}((1,1)^T)$$ while its reduced form $$R= \begin{pmatrix} 1&2\\0&0 \end{pmatrix}$$ would satisfy $$\text{Col}(R) = \text{span}((1,0)^T)$$
And in the other cases, namely $A$ is a $m \times n$ matrix with $m <n$ but the rank of $A$ is $<m$ and in the case that $m>n$ the result will also not be true - you can construct counterexamples pretty easily.