Is $(a)/(a^2)\cong R/(a)$, or something else? ; For ideals $(a), (a^2)\leq R$ (Rings are commutative with 1)
In a lecture on introductory commutative algebra, I was presented with an exercise that basically just asked: 

For a non zero-divisor $a\in R$, we have $(a)/(a^2)\cong$ ?

Well, a tempting guess would be $(a)/(a^2)\cong (1)/(a)\cong R/(a)$.
Trying to prove this, I just went on to construct the most natural isomorphisms I could think of. I first wrote down the proof, and later realized it had an error. I tried to fix it, but it seems whatever direction I choose, I always run into a similar problem like below; In showing $\psi$ being a ring homomorphism. I have left the proof as I first wrote it down, but with an adjustment pointing out the problem. (It's probably silly simple, it always tends to be when I stare too long at a problem.)
Question(s): Are there more problems with the proof below, and can I fix it; And if so, how?

Proof attempt. Let $\psi: (a)/(a^2)\to R/(a)$ be defined as $\psi(\bar{ar})=r+(a)=\bar{r}$.
Then, $\psi$ is well defined, since for some $r_1,r_2\in R$
$\overline{ar_1}=\overline{ar_2}\Rightarrow ar_1-ar_2=a(r_1-r_2)\in (a^2)\Rightarrow a(r_1-r_2)=a^2r_3=a(ar_3),$ for some $r_3\in R$
$\Rightarrow r_1-r_2=ar_3\Rightarrow r_1-r_2\in (a)\Rightarrow\bar{r_1}=\bar{r_2}\in R/(a)$.
Further, $\psi$ is a ring homomorphism since
$\psi(\overline{ar_1}+\overline{ar_2})=\psi(\overline{ar_1+ar_2})=\psi(\overline{a(r_1+r_2)})=\overline{r_1+r_2}=\bar{r_1}+\bar{r_2}=\psi(\overline{ar_1})+\psi(\overline{ar_2})$
$\psi(\overline{ar_1ar_2})=\psi(\overline{a^2r_1r_2)})\not=\bar{r_1}\bar{r_2}=\psi(\overline{ar_1})\psi(\overline{ar_2}) )$ $\rightarrow$ Problem!
In fact, as far as I can see, any product $\bar{ar_1}\bar{ar_2}$ in $(a)/(a^2)$ is $\bar{0}$.
Also, $\psi$ is surjective, since for any $\bar{r}\in R/(a)$, there is $\overline{ar}\in (a)/(a^2)$ s.t. $\psi(\overline{ar})=\bar{r}$ ;
And $\psi$ is injective since $\ker \psi = \{ar+(a^2)|r\in(a)\}=(a^2)=\bar{0}.$

Bonusquestion(s): Do ideals generated by one or more elements in a ring obey more canonical laws like this, such as $(a_1,a_2)/(a_1)=(a_2); (a_1,a_2)(b_1,b_2)=(a_1b_1,a_2b_2);$ and so on. Would someone refer me to some collected proofs of such if something like that exists. Can we give such ideals in a ring a ring structure on their own or something similar?
UPDATE: I have realized that we would run into the following problem, if my intuitive guess was correct:
By the third isomorphism theorem we have $(R/(a^2))/((a)/(a^2))\cong R/(a) \cong (a)/(a^2)$. This means we have a situation of $A/I=I$. This should not be nontrivially possible since $A$ unital implies $1+I$ is a unit in $A/I$, while $I$ contains no units if proper.
 A: Note that $(a)/(a^2)$ is not a ring (in general), but it is an $R$-module.
A good starting point is to consider the module homomorphism $f\colon R\to (a)/(a^2)$ defined by $f(x)=ax+(a^2)$. The map $f$ is surjective, so we can say that
$$
(a)/(a^2)\cong R/\!\ker f
$$
and it remains to determine $\ker f=\{x\in R:ax\in(a^2)\}$, which is usually denoted by $(a^2):a$.
This is an ideal of $R$ that contains $(a)$, but can be different. Indeed, if $a$ is a zero divisor and $ab=0$, with $b\ne0$, then $b\in\ker f$ but $b$ need not belong to $(a)$.
Note that your map $(a)/(a^2)\to R/(a)$ is generally not well-defined, because it only is when $\ker f=(a)$.
If $a$ is not a zero divisor, then $ax\in(a^2)$ implies $ax=a^2y$, for some $y$, and so $x=ay\in(a)$.
An example where $\ker f\supsetneq(a)$ is given by $R=\mathbb{Z}\times\mathbb{Z}$ and $a=(2,0)$. Then
$$
(a^2):a=\{(x,y)\in R: (x,y)(2,0)\in(a^2)\}
$$
and this is $2\mathbb{Z}\times\mathbb{Z}\ne(a)=2\mathbb{Z}\times\{0\}$.
Indeed, $(a)/(a^2)\cong\mathbb{Z}/2\mathbb{Z}$, but $R/(a)\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}$.
A: 
By the third isomorphism theorem we have $(R/(a^2))/((a)/(a^2))\cong R/(a)\cong(a)/(a^2)$. This means we have a situation of $A/I=I$. This should not be nontrivially possible since $A$ unital implies $1+I$ is a unit in $A/I$, while $I$ contains no units if proper.

I think a lot of the confusion stems from the fact that we consider morphisms of $R$-modules, not morphisms of rings. For example, if I have any element $a \in R$, then the $R$-linear homomorphism $\varphi:R \rightarrow (a), 1 \mapsto a$ is well-defines, because $R$ is a free module, so it suffices to define the homomorphism on a basis. Though we might consider the ideal $(a)$ as a ring, because it inherits a multiplication from $R$, this does not make $\varphi$ into a homomorphism of rings, because $\varphi(xy) = xya \neq xaya = \varphi(x)\varphi(y)$.  Also note, that $(a)$ does not need to have a multiplicatively neutral element, so this is not even a ring in your definition.
A quick counter-example is of course obtained by the zero-ring: $0/0 = 0$. A more elaborate counterexample is the so-called ring of dual numbers $R = k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field. This ring has a unique proper non-zero ideal $(\varepsilon)$. Then $R/(\varepsilon) \cong (\varepsilon)$, because both modules are (as $k$-vectorspaces) just $k$, and $\varepsilon$ acts as $0$ on both.
A necessary condition for $R/I \cong I$ is, that $I^2 = 0$, because $I$ acts as $0$ on $R/I$, so it has to act as $0$ on $I$ as well. This shows, that if $I = (a)$ is a principal ideal generated by a non-zerodivisor, this can never be that case.
Now to your problem: If $a\in R$ is a non-zerodivisor, the kernel of the map $\varphi$ is $\{0\}$, hence $\varphi$ is injective. Clearly it is also surjective, so we see that $R \cong (a)$, as $R$-modules. Now consider on both sides the submodule generated by multiplication with $a$, i.e. $a\cdot R$ and $a \cdot (a)$. As $\varphi$ is $R$-linear, it maps one bijectively to the other. Also note that $a \cdot R = (a)$ and $a \cdot (a) = (a)^2 = (a^2)$. Now $\varphi$ induces an isomorphism on the quotients.
$$\bar\varphi: R/(a) \xrightarrow{\cong} (a)/(a)^2.$$
