The answer to part (a) of your question is that the normal closure of $a^{k_1}\cdot t\cdot a^{k_2} \cdot t \cdot a^{k_3}$ is always equal to the normal closure of $a$ and $t^2$. This is because the quotient of $G$ obtained by adding a relation of the type written in part (a) always gives $a=1$!
Let me continue to work from now on in the quotient (and by abuse of notation with the same letters $a$ and $t$.) The added relation can be written in the form $t\cdot a^m\cdot t \cdot a^n=1$, so $t\cdot a^m = a^{-n} \cdot t^{-1}$. Then $b=t\cdot a\cdot t^{-1}=(t\cdot a^m)\cdot a\cdot (t\cdot a^m)^{-1}$, so also $b=(a^{-n} \cdot t^{-1})\cdot a\cdot (t\cdot a^n)$ and hence $t^{-1}\cdot a \cdot t = a^n \cdot b\cdot a^{-n}$.
It follows that not only does one have $b\cdot a\cdot b^{-1}=a^2$, but (after conjugating all letters in the last written relation through by $a^{-n}$) we can hence deduce $(t^{-1}\cdot a\cdot t)\cdot a\cdot (t^{-1}\cdot a\cdot t)^{-1}=a^2$. Now conjugating (every letter of) this last relation through by $t$ we obtain $a\cdot b\cdot a^{-1}=b^2$.
It is easy to see that $b\cdot a\cdot b^{-1}=a^2$ and $a\cdot b\cdot a^{-1}=b^2$ imply $a=1$.
The above combinatorial argument shows that in the one-relator group B (generators a,t with bab^{-1}=a^{2} and b=tat^{-1}) the normal closure of the word in part (a) contains a. A similar type of argument (using quotients) gives the same result for the word a^{k_{1}}ta^{k_{2}}t^{-1}a^{k_{3}}ta^{k_{4}} mentioned in part (b). Now B is also SQ universal (e.g. since it has G as a quotient.) I agree the problem starts to become more interesting with the words a^{k_{1}}ta^{k_{2}ta^{k_{3}ta^{k_{4}} of part (b). To begin with (and I hope this is an easier question) are there infinitely many non-isomorphic (perhaps one should require non-epimorphic) quotients of B using these words as relators?