It feels like a lot of shortcuts are being take here:
"The kernel is generated by the class of $a$ modulo $I$."
So, is the kernel just $(a+I)$?
"Since the kernel is generated by just one element, it has the form $R/J$ for some ideal $J$"
Is he using an isomorphism theorem here? If the kernel is $(a+I)$, then we have $\frac{R/I}{(a+I)/I} \cong R/(a+I)$?
"in fact, $J$ is the annihilator of $a$ modulo $I$"
Isn't the annihilator of $a$ modulo $I$ going to be $(I: a+I)$, not $(I:a)$?
First, there is a surjection $$ i\colon R/I\to R/(I+(a)):r+I\mapsto r+(I+(a)) $$ which is well-defined since $I\subseteq I+(a)$. Then $r+I\in\ker i$ iff $r\in (I+(a))$. Writing $r=x+sa$ for $x\in I$ and $s\in R$, $r+I=x+sa+I=sa+I$, so $r+I\in (a+I)$. Thus $\ker i\subseteq (a+I)$, and so $\ker i=(a+I)$ as the reverse containment is immediate. So yes, to your first question.
To say $\ker i=R/J$ for some $J$ is a bit iffy since $R/J$ is not necessarily a subring of $R/I$. One way to realize what this is saying is to note that there is a map $$ R\to R/I:r\mapsto ra+I $$ whose image is precisely $(a+I)$. Now $r$ is in the kernel iff $ra+I=I$, hence iff $ra\in I$, hence iff $r\in (I:a)$. So by the first isomorphism theorem, this map descends to an injective map $$ R/(I:a)\to R/I:r+(I:a)\mapsto ra+I $$ whose image is precisely $(a+I)=\ker i$. This is the multiplication by $a$ map, and gives the exact sequence.
For the third question, since $I$ is an ideal, a quick check shows $(I:a+I)=(I:a)$ as ideal quotients in $R$.