What is the meaning of $i^{\sigma ^{−1}}$? A string $x$ is a map $\Omega \rightarrow \Sigma$ from a finite set $\Omega$ of positions to a finite set $\Sigma$ of letters, the alphabet.
Let  $\Sigma^\Omega $ denote the set of all strings.
In examples we will write strings simply as chains of characters, e.g. $x = raspberry$ for $\Omega = \{1, \cdots 9\}$ and the lower case English alphabet$\Sigma= \{a, \cdots z\}$ .
The action of $\text{Sym}(\Omega)$ on $\Omega$ induces an action on $\Sigma^\Omega $.
For $ \sigma \in \text{Sym}(\Omega)$ and $x \in \Sigma^\Omega $, the string $x^\sigma $ is defined in this document (page 4) by:

$x^\sigma  (i) = x(i^{\sigma ^{−1}})$ for all $i \in \Omega$.

What is the meaning of $i^{\sigma ^{−1}}$?
Is $i$ the $i^{th}$ position in $x$? For example, if $x = raspberry$  then $x(2)=a$? Then if $\sigma = (12)$, $x^\sigma  (2)=arspberry$, right? then why the author used the inverse of $\sigma$ on $i$ in $x(i^{\sigma ^{−1}})$?
It would be natural to use $i^{\sigma}$ instead of $i^{\sigma ^{−1}}$, though author mentions, that "This twist is necessary in order to have .." I don't understand what author meant here.
 A: 
What is the meaning of $i^{\sigma^{-1}}$?

The paper defines $i^{\sigma^{-1}} = \sigma^{-1}(i)$.

Is $i$ the $i$th position in $x$?

We're told $i \in \Omega$, so $i$ itself has nothing to do with $x$ except that it ought to be in the domain of $x$.
In contrast, $x(i)$ is the image of $i$ under the map $x : \Omega \to \Sigma$, which in this particular case can be identified as the $i$th character of the string $x$.

For example, if $x=\mathrm{raspberry}$, then $x(2) = \mathrm{a}$?

Yes.

Then if $\sigma = (12)$, $x^\sigma(2) = \mathrm{arspberry}$, right?

No.  Rather, you have $x^\sigma = \mathrm{arspberry}$ and $x^\sigma(2) = \mathrm{r}$.

And now we finally reach your question: (in the future, it would be better to spotlight your actual question rather than your first question)

then why the author used the inverse of $\sigma$ on $i$ in $x(i^{\sigma^{-1}})$?

Short answer: We want to define $x^\sigma$ so that $(x^\sigma)^\tau = x^{\sigma\tau}$ and without the twist, we'd have $(x^\sigma)^\tau = x^{\tau\sigma}$ instead.  Therefore the twist is required.
Longer answer: This depends on how we define products of permutations in the symmetric group.  Unfortunately, the author of the paper didn't deem it necessary to explicitly mention how to evaluate products in the preliminaries section, but they did give us their reference for "a comprehensive introduction" to permutation groups: Dixon and Mortimer's Permutation Groups.  Looking it up, we find this on page 3:

Our convention is to consider permutations as functions acting on the right. This means that a product $xy$ of permutations should be read as: first apply $x$ and then $y$

In other words, in the current context, $(\sigma\tau)(i) = (\tau\circ \sigma)(i) = \tau(\sigma(i))$, which might feel backwards to you and is probably why the author of the paper prefers the alternative notation $i^\sigma = \sigma(i)$, as under this notation,
$$i^{\sigma\tau} = (\sigma\tau)(i) = \tau(\sigma(i)) = \tau(i^\sigma) = (i^\sigma)^\tau.$$
That is, $i^{\sigma\tau} = (i^\sigma)^\tau$, which is much better!  Anyway, this choice has the effect of forcing us to use a twist when defining $x^\sigma$.
If instead we defined $x^\sigma$ without the twist as $x^\sigma(i) = x(i^\sigma)$: Then we'd have
$$x^{\sigma\tau}(i) = x(i^{\sigma\tau}) = x((i^\sigma)^\tau) = x^\tau(i^\sigma) = (x^\tau)^\sigma(i)\\ \implies x^{\sigma\tau} = (x^\tau)^\sigma$$
Note that the order has been "messed up" without the twist.
When defining $x^\sigma$ with the twist as $x^\sigma(i) = x(i^{\sigma^{-1}})$:  Then we have
$$x^{\sigma\tau}(i) = x\left(i^{(\sigma\tau)^{-1}}\right) = x\left(i^{\tau^{-1}\sigma^{-1}}\right) = x\left(\left(i^{\tau^{-1}}\right)^{\sigma^{-1}}\right) = x^\sigma\left(i^{\tau^{-1}}\right) = (x^\sigma)^\tau(i) \\ \implies x^{\sigma\tau} = (x^\sigma)^\tau$$
and the order is nice, once more!

A couple final remarks:

*

*Note that the paper has a typo in the middle of equation (2.1) when showing that the twist yields the preferred result of $x^{\sigma\tau} = (x^\sigma)^\tau$.  I've corrected it in my version I've shown here.

*Even if you chose the convention that permutations act on the left, you'd probably still want to use a twist when defining the action of a permutation $\sigma$ on the function $x$.  This is because the twist preserves both left actions and right actions, while the untwisted version turns left actions into right actions and vice versa.

A: I believe it is the image of $i$ under $\sigma^{-1}$.
