What is the relationship between the null space and the column space?

You're close. What you are touching on is the fundamental theorem of linear algebra, which establishes that there are what is known as the $4$ fundamental subspaces of an $m\times{n}$ matrix.

The row space and the nullspace together span the domain of the linear transformation: $\mathbb{R}^n$. Their intersection only contains $1$ element: the $n$ component $0$ vector.

Similarly, the column space and the left nullspace together span the co-domain of the linear transformation: $\mathbb{R}^m$. Their intersection only contains $1$ element: the $m$ component $0$ vector.

Furthermore, the rank of the matrix is the dimension of both the column space and the row space. The dimension of the nullspace is $n-r$, and the dimension of the left nullspace is $m-r$.

Your wording is a little unusual: the null space of those vectors. The null space is usually defined for a linear function, not for a set of vectors. You can, nonetheless, put those vectors together and make up a linear function. More specifically, define $T:\mathbb R^m \to \mathbb R^n$ by $$ T(\vec a) = a_1\vec v_1 + a_2\vec v_2 + \ldots + a_m\vec v_m. $$ where $a_i$ are components of $\vec a$. Or in matrix-vector form, this is $$ T\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \begin{pmatrix} | & | & \dots & | \\ \vec v_1 & \vec v_2 & \ldots & \vec v_m \\ | & | & \ldots & | \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \sum_{i=1}^m a_i\vec v_i. $$ Then you can talk about the null space of $T$. The space spanned by $\vec v_i$ is the image of $T$, which is exactly the column space of the matrix $(\vec v_1\ \ \vec v_2\ \ \ldots\ \ \vec v_m)$. Now, the rank-nullity theorem coincides with what you stated.

A little more explanation: The vector $\vec a \in \mathbb R^m$ indicates a choice of linear combination for a fixed set of vectors $\vec v_i$. $\vec a$ will belong to the null space of $T$ if $T(\vec a) = 0$, i.e., $\vec a$ corresponds to a linear combination that is zero. If $\vec v_i$ are linearly independent, the only linear combination that is zero is the zero linear combination, which is when all coefficients are zero. Then the null space of $T$ has only one member: $\vec a = 0$. In a sense, the dimension of the null space of $T$ measures how dependent (or redundant) $\vec v_i$ are.