Writing this up led me to the solution.
When we plug in values into the pmf $f(x,y)$ such as $f(1,2)$ we must recall that we have already decided what the value of $X$ is. Although there are two ways to sum two four-sided to get the value $2$, we have already set $X=1$ and therefore there is only one value ($1$) that the remaining die may have to make $2$. Thus there is only $1$ way to create $2$ given the parameters of the pmf and thus the probability for this and all rolls is static.
$f(x,y) = \frac{1}{16}$
