$(3\Bbb Z/6\Bbb Z) \cong (\Bbb Z/2\Bbb Z)$ can be easily shown by using the first isomorphism theorem, but I heard that we cannot say that $(2\Bbb Z/6\Bbb Z) \cong (\Bbb Z/3\Bbb Z)$.
Why not? And what are the conditions in which we can "simplify" the ring?
The confusion may rely on your definition of $\cong$.
If $\cong$ means group isomorphism then the formula $(2\mathbb Z/6\mathbb Z) \cong (\mathbb Z/3\mathbb Z)$ is true.
If the use of $\cong$ here means canonically isomorphic then the formula is false.
Recall that for a given class of algebraic structures (say rings, groups, fields, ...) two objects in the class are said to be canonically isomorphic if there is a unique isomorphism between them.
In your examples $3\mathbb Z/6\mathbb Z$ is a group of two elements and admits a unique isomorphism with $\mathbb Z/2\mathbb Z$, so they are canonically isomorphic. Indeed, in an isomorphism $$ 3\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/2\mathbb Z $$ the class of $0$ needs to be mapped to the class of $0$. This leaves no freedom of choice for the image of $[3]$.
However, in the case $2\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/3\mathbb Z$ there are $2$ isomorphisms (of groups) completely determined by the image of $2$. In this case any chocie $[2]\mapsto [1]$ or $[2]\mapsto [2]$ defines an isomorphism.
Further remarks Let $m,k$ be positive integers. Then $k \mathbb Z/mk\mathbb Z$ is a cyclic group of order $m$. Thus, it is isomorphic to $\mathbb Z/m\mathbb Z$ as groups. The number of possible isomorphisms $k \mathbb Z/mk\mathbb Z\rightarrow \mathbb Z/m\mathbb Z$ coincides with the number $\varphi(m)$ of generators of $\mathbb Z/m\mathbb Z$. Here $\varphi(m)$ denotes the Euler's totient function which is $\geq 2$ if $m \geq 3$.
Hence $k \mathbb Z/mk\mathbb Z\cong \mathbb Z/m\mathbb Z$ is always true as group isomorphism. However $k \mathbb Z/mk\mathbb Z$, $\mathbb Z/m\mathbb Z$ are only canonically isomorphic when $m=1$ or $2$.
EDIT:
Notice that the canonically part depends strongly on the class of structures you consider, as the comments suggest. The ring $2\mathbb Z/6\mathbb Z$ is unital (the unit is $4$) and $2\mathbb Z/6\mathbb Z$ is canonically isomorphic to $\mathbb Z/3\mathbb Z$ as unital rings.