A difficulty in understanding an example in Hungerford Algebra. The example is given in the following picture:

Here is the text for clarity

EXAMPLES. If $I$ is a left ideal of a ring $R$, then $I$ is a left $R$-module with $ra(r \varepsilon R,a \varepsilon I)$ being the ordinary product in $R$. In particular, $0$ and $R$ are $R$-modules. Furthermore since $I$ is an additive subgroup of $R$, $r / I$ is an (abelian) group. $R/I$ is an $R$-module with $r(r_1+I) = rr_1 + I$. $R/I$ need not be a ring however unless $I$ is a two sided ideal.

It is not clear for me how "$I$ is a left $R$-module with ra being the ordinary product in $R$", I know that by the definition of an ideal it is a ring and hence an additive abelian group but how the ring action is applied it is not clear for me the details, how multiplication of a ring element is distributed over addition of 2 ideal elements and so on, could anyone clarify this for me please? 
Also,I did not understand why "$R/I$ need not be a ring, however, unless $I$ is a two-sided ideal" could anyone explain this for me please?
 A: The meaning of the sentence seems straightforward so I guess you are looking for a counterexample.
Let $T$ be the right ideal of matrices in $M_2(F)$ with bottom row zero.
$T$ is not closed by multiplication on the left by $A=\begin{bmatrix}0&0\\ 1&0\end{bmatrix}$.
Then multiplication on the cosets isn't  well defined since  $B=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}\equiv 0$ mod $T$,  but $AB\not\equiv 0$ mod $T$.

It is not clear for me how "$I$ is a left $R$-module with $ra$ being the ordinary product in $R$"

The module structure on $R/I$ is given by $r\cdot(a+I):=ra+I$. This is well-defined since $I$ is a left $R$ module. That is, if $r=r'$ and $a+I=a'+I$, we first have that $a-a'\in I$, and therefore $r(a-a')\in I$. This says $ra+I=ra'+I$. Since multiplication in $R$ is well-defined, $ra'=r'a'$ and hence $ra'+I=r'a'+I$. Stringing things together, $ra+I=r'a'+I$. Therefore the map $R\times R/I\to R/I$ is well-defined.
It is an easy matter to show that the rest of the module axioms hold for $R/I$ because they hold for $R$ itself.
