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 $a=1$.
P.S.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 after part (b). Now $B$ is also SQ universal (e.g. since it has $G$ as a quotient.) I agree with Harald 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 may be an easier question) are there infinitely many non-isomorphic quotients of $B$ using these words as relators?