Let say we have $$ \frac{\partial |F|}{\partial t} \le K |F| + K \alpha^{2},\:\:\:\:\: t \in [t_{n}, t_{n+1}]$$ with $F(x, t_{n})=0$ and $t_{n+1}-t_{n}=\triangle t$. The above is the same thing as $$ \frac{\partial (|F| + \alpha^{2})}{\partial t} \le K ( |F| + \alpha^{2})$$
and by Gronwall's inequality we must have that $|F|$ is bounded by the solution of $$\frac{\partial (|f| + \alpha^{2})}{\partial t} = K ( |f| + \alpha^{2})$$
which is $$ |f|+\alpha^{2} = g(x)e^{Kt}$$
but $|f(x,t_{n})|=0$ which means $\alpha^{2} = g(x)e^{Kt_{n}}$. So by Gronwall's inequality we have
$$ |F| \le \alpha^{2} e^{K(t-t_{n})} - \alpha^{2}$$
but why does the author of a paper mentioned $$ |F| \le K_{\alpha} \triangle t $$ where $K_{\alpha}$ is some constant depending on $\alpha>0$?
I can only guess without more information, but I suppose that only small time intervals $\Delta t$ are considered. Indeed, if $\Delta t\leq C$, then the right hand side of the second last equation of your question can be rewritten as follows: $$\alpha^2\left(e^{K(t-t_n)}-1\right)=\alpha^2\left(\sum_{k=1}^\infty \frac{K^k(t-t_n)^k}{k!}\right)=\alpha^2K(t-t_n)\left(\sum_{k=0}^\infty \frac{K^{k}(t-t_n)^k}{(k+1)!}\right)\leq\alpha^2K\Delta t\, e^{CK}.$$