I'm going to ignore the \$V_{_\text{THN}}\$ term and focus on the remaining term, which is a product of two functions.
If you have the product of two functions, call them \$P\$ and \$Q\$, so that \$F=Q\cdot P\$, then the product rule would have:
$$\begin{align*}\text{d} F&= P\cdot\text{d} Q+Q\cdot\text{d} P\\\\&= \frac{Q}{Q}\cdot P\cdot\text{d} Q+\frac{P}{P}\cdot Q\cdot\text{d} P\\\\&= Q\cdot P\cdot\frac{\text{d} Q}{Q}+P\cdot Q\cdot\frac{\text{d} P}{P}\\\\&= Q\cdot P\cdot\left[\frac{\text{d} Q}{Q}+\frac{\text{d} P}{P}\right]\tag{1}\label{eq:1}\end{align*}$$
The way to read the above Eq. \ref{eq:1} is to think: "The differential change in the product of \$Q\$ and \$P\$ is that same product times the sum of the %-change in \$Q\$ plus the %-change in \$P\$."
(Note that I've not yet actually taken the derivative with respect to any other infinitesimal. In fact, I don't yet even know what \$P\$ and \$Q\$ are functions of. And I don't care. This is how I prefer to perform calculus. I don't know anything about \$P\$ or \$Q\$ except that they may yet become functions of something. Yet I have already been able to say something useful about the differential of their product!)
In your specific case:
\$P\$ is a function of \$K\$, so write: \$P_{_\text{K}}=1-\frac1{\sqrt{K}}\$.
\$Q\$ is a function of \$R\$, so write: \$Q_{_\text{R}}=\frac2{\beta_1\,R}\$.
$$\begin{align*}\require{cancel}\text{d} F&= Q\cdot P\cdot\left[\frac{\text{d} Q}{Q}+\frac{\text{d} P}{P}\right]\\\\\frac{\text{d} F}{\text{d}\,T}&= Q_{_\text{R}}\cdot P_{_\text{K}}\cdot\left[\frac{\frac{\text{d}\, Q_{_\text{R}}}{\text{d}R}}{Q_{_\text{R}}}\cdot\frac{\text{d}\,R}{\text{d}T}+\frac{\frac{\text{d}\, P_{_\text{K}}}{\cancel{\text{d}K}}}{P_{_\text{K}}}\cdot\frac{\cancel{\text{d}\,K}}{\text{d}T}\right]\\\\&\quad\quad\text{substituting, } \frac{\frac{\text{d}\, Q_{_\text{R}}}{\text{d}R}}{Q_{_\text{R}}}=-\frac1{R}\\\\&= Q_{_\text{R}}\cdot P_{_\text{K}}\cdot\left[\frac{-\text{d}\,R}{R}\cdot\frac{1}{\text{d}T}+\frac{\text{d}\, P_{_\text{K}}}{P_{_\text{K}}}\cdot\frac{1}{\text{d}T}\right]\tag{2}\label{eq:2}\\\\&= Q_{_\text{R}}\cdot P_{_\text{K}}\cdot\left[\frac{-\text{d}\,R}{R}+\frac{\text{d}\, P_{_\text{K}}}{P_{_\text{K}}}\right]\cdot\frac{1}{\text{d}T}\tag{3}\label{eq:3}\\\\&\quad\quad\text{or also,}\\\\\frac{\text{d} F}{F}&=\frac{\text{d}\, P_{_\text{K}}}{P_{_\text{K}}}-\frac{\text{d}\,R}{R}\tag{4}\label{eq:4}\end{align*}$$
That's what I come up with just applying math and not attempting to read the paper with understanding.
Take note that Eq. \$\ref{eq:2}\$ casts, yet again, things in terms of %-change in \$R\$ and %-change in \$P_{_\text{K}}\$. Looking at things as %-change helps focus on a sensitivity viewpoint, which is a very powerful way to consider issues at hand. To ask, "How sensitive is the output to a %-change in input \$x\$ or a %-change in input \$y\$?", is quite powerful!
Compare Eq. \$\ref{eq:2}\$ with equation 7 in your document, perhaps. I suspect their KP(T) is \$P_{_\text{K}}\$, above, knowing that K is a function of T. (Eq. \$\ref{eq:3}\$ is just another way of writing it.)
I think Eq. \$\ref{eq:4}\$ may also be important to understand. It says how a %-change in \$V_{_\text{REF}}\$ depends on %-changes elsewhere. And it suggests a way to consider how one may attempt to zero-out, or at least minimize the worst case of, any variation in \$V_{_\text{REF}}\$ by balancing two terms. But it isn't covered in your document. So I'll leave it, for now.