Intro
a right R-module is equivalent to a left R-module only when R is commutative
This statement has a pair of serious problems to resolve.
Because "equivalent" is undefined, it's unclear what the statement means. (Discussed briefly below.)
It uses "only when," but that is the wrong logical direction: it should use just when. There are in fact not-commutative rings where left and right modules are pretty much the same, so "only when" is not really appropriate.
I think a better version of the statement is:
A right $R$ module structure can be used as a left $R$ module structure when $R$ is commutative
Modules over commutative rings
Suppose you've defined a right $R$ module for a commutative ring so that $mr$ makes sense for $m\in M$ and $r\in R$. Then naively one might say "oh, well that must be the same thing as $rm$."
Actually, it's not immediately clear that that is a legitimate thing to do, but yes, using commutativity of $R$, you can verify that the new action $rm:=mr$ satisfies all the module axioms. After that is done, we would have used the right $R$ module structure as a left $R$ module structure.
Symmetrically, left module structures can be used as right module structures, and in fact if you switch sides twice, you wind up with the original module you started with. That's why we say when $R$ is commutative, right and left modules are "the same."
Modules over rings in general
It turns out that a left $R$ module structure on an abelian group $M$ amounts to a ring homomorphism of $R\to End(M)$, where $End(M)$ is the set of group endomorphisms of $M$.
On the other hand, a right $R$ module structure on $M$ amounts to a ring homomorphism from $R\to End(M)^{op}$, where $End(M)^{op}$ is the opposite ring of $End(M)$. (Equivalently, you could use a ring homomorphism from $R^{op}\to End(M)$.)
This whole "opposite ring" business is what makes it necessary to keep track of module sides: the homomorphisms of $R$ into $End(R)$ might be completely different from those into $End(R)^{op}$. That's just the way things are.
Finally, if $R\cong R^{op}$, then something nice happens! Since $R$ and $R^{op}$ can't be distinguished, you can in fact use a left $R$ module structure as a right $R$ module structure. Formally, if you had the homomorphism $R\to End(M)$, you could simply compose it with the isomorphism $R^{op}\to R\to End(M)$ obtaining $R^{op}\to End(M)$, a right $R$ module structure.
This is obviously the case for commutative rings, since the identity map is an isomorphism of a commutative ring with its opposite ring. But this applies more generally, since there are not-commutative rings isomorphic with their opposite rings.
Your second question
Your question is based on good observations :) The previous paragraph supplies a partial explanation of why the internal property of commutative rings can influence the two categories of modules. The reason is that since commutativity makes the ring "symmetric," its right modules and left modules are going to look alike.
If you want to deepen your understanding of how the properties of a ring and its two module categories interact, then you will have a good time studying module theory because that is one of module theory's main topics.
Best Answer
Over a commutative ring there is no difference between a left and right module. In general the difference between a left and right module is that in a left module $$a\cdot(b\cdot m)=(ab)\cdot m$$ while in a right module $$(m\cdot a)\cdot b=m\cdot(ab)$$
So if a ring $R$ is non commutative there is a difference between a left and right action. Given a left action $a\cdot m$ if one tries to define $m\cdot a=a\cdot m$ this does not in general give a right action, because $$(m\cdot a)\cdot b=b\cdot(m\cdot a) = b\cdot(a\cdot m)=(ba)\cdot m$$ while $$m\cdot (ab)= (ab)\cdot m$$ However, if $R$ is commutative then there is no problem. Every left action defines a right action which is "the same" and vice versa. In this case it is a matter of convenience whether one chooses to write $a\cdot m$ or $m\cdot a$.