From Algebra: Chapter $0$ by Aluffi:
Since $f(x) \in k[x]$ is irreducible, any irreducible factor must be over the splitting field of $f(x)$.
So, $q(x)$ above must be a polynomial over a splitting field.
Being an associate is dependent on which ring we are working in. For example, over $\mathbb Z$, the only associates of $2$ are $2$ and $-2$; but over $\mathbb Q$, every non-zero element is an associate.
I don't understand how $q$ and $f$ must be associates if over the splitting field, $f(x)$ is no longer irreducible.
If $p(x)$ and $q(x)$ are both in $k[x]$, then their greatest common divisor in $k$ is the same as the greatest common divisor in $\overline{k}[x]$. Because $k[x]$ is a Euclidean domain.
If $g(x)$ is the greatest common divisor over $k[x]$ and $f(x)$ is the greatest common divisor in $\overline{k}[x]$, then certainly $g(x)$ divides $f(x)$. But since we can express $g(x)$ as $$g(x) = a(x)p(x) + b(x)q(x)$$ with $a(x),b(x)\in k[x]$, then since $f(x)$ divides both $p(x)$ and $q(x)$, it also divides $a(x)p(x)+b(x)q(x)=g(x)$. Hence $f(x)$ divides $g(x)$, so $f(x)$ and $g(x)$ are associates in $\overline{k}[x]$.
But they are both monic; so they are actually equal. Thus, $g(x)$ is also the gcd in $\overline{k}[x]$.
Now, since $f(x)$ is monic and irreducible in $k[x]$, its gcd with any polynomial in $k[x]$ is either $f(x)$ or $1$. Since $f(x)$ is inseparable, $\gcd(f,f')\neq 1$. So $\gcd(f,f')=f$. So $f|f'$. But since $f'$ is either zero or of degree strictly smaller than $f$, the only way $f$ can divide $f'$ is if $f'=0$. So $f'=0$.