Artin Theorem 14.9.1
Let $V$ be a finitely generated module over the polynomial ring $\mathbb{C}[x_1,...,x_k]$, and let $A$ be an $m\times n$ presentation matrix for $V$. Denote by $A(c)$ the evaluation of $A$ at a point $c$ of $\mathbb{C}^k$. Then $V$ is a free module of rank $r$ if and only if the matrix $A(c)$ has rank $m-r$ at every point $c$.
I want to use the above theorem from Artin's algebra book, but alas, I am not sure if I understand it correctly and thus I will be skeptical of any result I prove with it.
By saying "$V$ is a free module of rank $r$", I am pretty sure Artin means $V\cong (\mathbb{C[x_1,...,x_n]})^r$. By saying "$A(c)$ has rank $m-r$", I my guess is that he means dim(colspace($A(c)$)) $=m-r$, where we understand colspace$(A(c))\subseteq \mathbb{C}^m$ to be a vector space over the field $\mathbb{C}$.
However, I am not entirely sure that I am interpreting the theorem correctly. Hopefully, someone can point me in the right direction. Ideally, I would like an answer linking to a similar theorem from another source (that laypeople can understand), so I don't have to trust Artin.
You are interpreting the word "rank" correctly, as it is used in two different ways, although $A(c)$ is an $n\times m$ matrix over $\mathbb C$, as you have taken the entries of the matrix $A$ and evaluated each of them at a point, going from a matrix of polynomials to a matrix of numbers. However, let's try to unravel a bit of what is going on so that the proposition makes sense.
Evaluation of an element of $R=\mathbb C[x_1,\ldots, x_k]$ at a point $(a_1,\ldots, a_n)$ is the same as taking its equivalence class in the quotient $R/\mathfrak m$ where $\mathfrak m$ is the maximal ideal $(x_1-a_1,x_2-a_2,\ldots, x_k-a_k)$. By the Nullstellensatz, every maximal ideal is of this form, and so you don't need to worry that evaluation is something special that you only get to do in this particular example, as the ideas generalize.
Given an ideal $I$ of a ring $R$, quotienting by $I$ gives you a ring $R/I$, but it gives you more. Given a module, you get a quotient module $M/IM$ (which is a module over $R/I$), and given a map of modules $M\to N$, you get an induced map of $R/I$ modules $M/IM\to N/IN$.
Given a finite presentation (which we could write as a matrix, but I will instead write with maps) $R^n\to R^m \to R^m/\operatorname{im}(R^n)\cong M$, we can look at the induced map $(R/I)^n\to (R/I)^m \to (R/I)^m/\operatorname{im}(R/I^n)\cong M/IM$, where we have used the fact that $R^n/I(R^n)\cong (R/I)^n$. I don't know how much you have seen of this, but it may be worth at least trying to prove that if a map is surjective, then the induced map on the quotients is still surjective.
Now, let's specialize to the case in the theorem, where $R$ is a polynomial ring, our ideal that we are quotienting by is maximal, and therefore the quotient ring is $\mathbb C$, and so finitely generated modules are just vector spaces and we can just do regular linear algebra. If we had a presentation of a module of rank $r$, then quotienting yields $\mathbb C^n\to \mathbb C^m$ (given by a $m\times n$ matrix), and the quotient of $\mathbb C^m$ by the image of the map (which is a vector space of dimension the rank of the map) is isomorphic to $\mathbb C^r$. Since the quotient of a vector space of dimension $a$ by a vector subspace of dimension $b$ is a vector space of dimension $a-b$, this means that rank of the (evaluated) presentation matrix has to be $m-r$. This actually proves one direction of the theorem. But more importantly, it explains where the numbers in the theorem come from and why someone might stumble upon the proposition to begin: if you have a situation you want to understand, try quotienting by maximal ideals and see what linear algebra over a field tells you.
The other direction is slightly less straight forward, and doesn't follow immediately from just unraveling what the theorem is saying. A priori, the ranks of the reduced map could be the same without the quotient being free, and there are more modules to consider than just free ones. However, since the proof is presumably in the book, there is no need for me to put one here.