Assume that we are given a morphism $m: X\to Y$ of varieties and that $I\subset O_Y$ is an ideal sheaf defining some subscheme $T\subset Y$. Then we have two objects on $X$ associated to $I$. The one is $m^{-1}I\cdot O_X$, the other is $m^*I=m^{-1}I\otimes O_X$. From the inclusion $i: I\to O_Y$ we get $m^*i: m^*I\to m^*O_Y=O_X$ and I found the statement that $m^{-1}I\cdot O_X=(m^*i)(m^*I)$.
Now there are some questions occurring to me.
Relation between the subscheme defined by $T$ and those defined by $m^{-1}I\cdot O_X.$
Let me denote the subscheme of $X$ defined by $m^{-1}I\cdot O_X$ by $S$. Now, if $I_T=\langle f_1,\dots,f_n\rangle$, is $I_S=\langle m^*f_1,\dots, m^*f_n\rangle$? It seems that this should rather be $m^*I_T$ but how does $m^{-1}I\cdot O_X$ then look like?
Why is $m^*I$ not necessarily a subsheaf of $O_X$?
I know the short answer is that it is because of non-exactness of the tensor product but its hard for me to find an explicit example. This maybe has to do with me not understanding the previous question to well, i.e. not understanding how $m^{-1}I\cdot O_X$ looks like.
If $m^{-1}T=S$, does $m^{-1}I_T\cdot O_X=I_S$ follow?
By $I_S$ I mean the ideal sheaf defining $S$ and by $I_T$ I mean the ideal sheaf defining $T=m^{-1}S$. Since it is an ideal sheaf, the preimage of the ideal defining $T$ would then be the ideal defining $T$ and it looks like this should then give the statement on the ideal sheaves but I am unable to write down a proof.
As you said, in general, $m^* I$ is not an ideal sheaf of $O_X$-modules, since the pullback functor is only right-exact, but there is an exact sequence:
$m^* I \rightarrow O_X \rightarrow m^* O_T \rightarrow 0$
and $m^{-1}I \cdot O_X$ is equal to the ideal sheaf generated by the image of $m^*I$ in $O_X$.
I think that the ideal $I_S$ you wrote down in the first question is actually $m^{-1}I\cdot O_X$ (where $m^*: O_Y \rightarrow m_* O_X$ is the canonical map of sheaves associated to $m$), while $m^*I$ has a more abstract definition.
For the last question, if $m^{-1}T$ is obtained by pullback of schemes, then its defining ideal is $m^{-1}I_T \cdot O_X$, hence your statement is true.
You can find all this and the example in Vakil's AG 16.3.9.