The rank $r$ of a matrix $A \in \mathbb{R}^{m \times n}$, as you have said is the dimension of the column space ($r$ is also the dimension of the row space as well) i.e. the dimension of the space spanned by vectors which are obtained by a linear combination of the columns of $A$, equivalently the range of $A$. (The use of the word "minimum" in the question is unnecessary). However each column vector has $m$ components and the vectors in the range of $A$ has $m$ components as such but span only a $r (\leq m)$ dimensional subspace instead of a $m$ dimensional space. So we are missing out spanning the remaining $m-r$ dimensional subspace of the $m$ dimensional space.

The left null-space now plays the roll of spanning the remaining $m-r$ dimensional subspace. This is why the left null-space is orthogonal to the column space. So the left null-space along with the column space now spans the entire $m$ dimensional space i.e. if $C = \{y \in \mathbb{R}^{m \times 1}: y = Ax\text{ for some }x \in \mathbb{R}^{n \times 1} \}$ and $Z_L = \{z \in \mathbb{R}^{m \times 1}:z^T A = 0 \}$,

then $Z_L \cup C = \mathbb{R}^{m}$ and $Z_L \perp C$

The right null-space plays the analogous roll for the rows. The rows span only a $r$ dimensional subspace of the $n$ dimensional space. The right null-space now plays the roll of spanning the remaining $n-r$ dimensional subspace. This is why the right null-space is orthogonal to the row space. So the right null-space along with the row space now spans the entire $n$ dimensional space i.e. if $R = \{y \in \mathbb{R}^{n \times 1}: y = A^Tx\text{ for some }x \in \mathbb{R}^{m \times 1} \}$ and $Z_R = \{z \in \mathbb{R}^{n \times 1}: Az = 0 \}$,

then $Z_R \cup R = \mathbb{R}^{n}$ and $Z_R \perp R$