On page 29 of this book order of ideal sheaf is defined
Let X be a smooth variety and $0 \neq I \subset \mathcal{O}_X$ be an ideal sheaf. For a point $x \in X$ with ideal sheaf $m_x$, we define the order of sheaf to be $$\operatorname{ord}_x I := \operatorname{max}\{ r: m_x^r\mathcal{O}_{x,X} \subset I\mathcal{O}_{x,X}\}$$
My confusion lies in the notation $I\mathcal{O}_{x,X}$ and $m_x^r\mathcal{O}_{x,X}$. Can you someone explain me what does $I\mathcal{O}_{x,X}$ mean?
Also,
$$\operatorname{max-ord}_X(I)= \operatorname{max}\{\operatorname{ord}_x I: x \in X\}$$.
If $\operatorname{max-ord}_X(I)=0$, why is I locally principal?
Firstly, your definition is the wrong way around: it should be $m_x^r\mathcal{O}_{x,X}\supset I\mathcal{O}_{x,X}$. (Use '\supset', not '\subset', as the latex command.)
Secondly, the notation $AR$ for a subset $A$ in a ring $R$ is another way of expressing the ideal generated by $A$.
Finally, if $\operatorname{max-ord}_X(I)=0$, this implies that $X$ has no points at which $I\mathcal{O}_{x,X}\subset m_x\mathcal{O}_{x,X}$. Suppose $I_x$ was generated as an $\mathcal{O}_{x,X}$ module by at least two elements $s_1,s_2,\cdots,x_n$ for some point $x$. For $I_x\not\subset m_x\mathcal{O}_{x,X}$, at least one of these (we'll assume up to reindexing it is $s_1$) must be in $I_x\setminus m_x$. This means that $s_1$ is a unit and by action of $\mathcal{O}_{x,X}$ we may obtain $s_2,\cdots,s_n$ by acting appropriately on $s_1$. Therefore $I_x$ is generated by one element for all $x$, and therefore $I$ is locally principal.