Here are my comments combined into an answer.

For ascending HNN extensions, i.e. $H=K$, $\phi\colon H\to K'$ an injective endomorphism (as in Baumslag-Remeslennikov case (see above), in the Grigorchuk case, and many others) one needs, as Ben Steinberg ponted out that $H$ has a finite L-presentation (named after Igor Lysenok, who proved that the Grigorchuk group $G_1$ has such a presentation) with respect to the endomorphism $\phi:H\to K′$. That is there are finite number of relations $r_1=1,...,r_k=1$ so that the set of relations $\{\phi^m(r_j)=1\mid m\ge 0,1\le j\le k\}$ defines $H$. Here we consider $\phi$ as a substitution $x\mapsto u_x$ where $u_x$ is any word representing $\phi(x)$ in $H$, $x$ a generator of $H$. Indeed, in this case $G$ (generated by the finite generating set $X$ of $H$ and the free letter $t$) has finite presentation consisting of relations $r_1,...,r_k$ and the HNN relations $x^t=\phi(x), x\in X$. I think that the converse statement should also be true: if the HNN extension is finitely presented then $H$ has a finite L-presentation with respect to $\phi$.

More examples can be found in Sapir, Mark, Wise, Daniel T., Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202 and in Olʹshanskii, Alexander Yu.; Sapir, Mark V. Non-amenable finitely presented torsion-by-cyclic groups. Publ. Math. Inst. Hautes Études Sci. No. 96 (2002), 43–169. In both cases it was crucial that the "extended" group $H$ has an L-presentation, in fact it was constructed as such.

For arbitrary HNN extensions the situation is more difficult but not hopeless, I think that necessary and sufficient conditions can be found in many more cases.

theserestrictions on $G$ and $K$ (and perhaps on the isomorphism between $K$ and $K^{\prime}$) then you can say something." Along the lines of the Baumslag-Tretfoff conditions for residual finiteness; sufficient, but not necessary. – user6503 Aug 10 '12 at 12:05