Is $(a)/(a^2)\cong R/(a)$, or something else? ; For ideals $(a), (a^2)\leq R$

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 $R/(a)\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}$.