Normal distribution parametrization I have the following hierarchical model:
$y_{i} = \alpha + \beta_{i}x_{i} + \varepsilon_{i}     $    where $\varepsilon \sim N(0,\sigma^{2})$.
$\beta_{i} \sim N(\gamma x_{i},\sigma^{2}x_{i}^{-2})$
Knowing this information I would like to understand why we can redifine the prior of $\beta_{i}$ as following:
$\beta_{i} = \gamma x_{i} + \sqrt{\sigma^{2}x_{i}^{-2}}\eta_{i}$ with $\eta_{i} \sim N(0,1)$.
I don't understand the intuition and the mathematical property/derivation that allows us to express it as the mean plus square root of the variance. Some clarification on this would be really helpful!
 A: Your last line could also have been written like this:
$$
\beta_i = \gamma x_i+\zeta_i
$$
with:
$$
\zeta_i\sim N\left(0,\sigma^2x_i^{-2}\right)
$$
Any variable with a scaled Normal distribution can be rewritten as a product between a variable with a standard (i.e. unit variance) Normal distribution, and some scaling factor equal to the standard deviation of the desired scaled Normal. In your case $\sqrt{\sigma^2x_i^{-2}}$ is the standard deviation of the scaled Normal that describes the variability in $\beta_i$, such that:
$$
\zeta_i=\eta_i\sqrt{\sigma^2x_i^{-2}}
$$
If we substitute this into the equation at the top of my answer, you get back the equation in your question.
A: Regarding the lines $$\beta_i\sim\dots$$ to $$\beta_i=\dots$$ it is just about the normalization of the normal distribution (note that variance is square of standard deviation). 
A: As I see it, your question is why a random variable $X \sim N(\mu, \sigma^2)$ is equivalent to $Y = \mu + \sigma Z$ with $Z \sim N(0,1)$.
A very informal explanation is as follows. Consider a standard normal random variable $Z \sim N(0,1)$ and then let $Y = \mu + \sigma Z$ (with $\sigma > 0$). If you think about the Bell curve, you'll see that the transformation of multiplying by $\sigma$ and adding $\mu$, then this keeps the Bell curve intact, it just moves it so that its center is at $\mu$ and expands it such that its width is inflated (or shrunk, for $\sigma < 1$) by $\sigma$. Now all we need to do is make sure that the mean and variance of $Y$ are as expected. As you see from the Wikipedia links below, these are standard properties of the mean and variance.
https://en.wikipedia.org/wiki/Expected_value#Properties
https://en.wikipedia.org/wiki/Variance#Properties
We have
$E(Y) = E(\mu + \sigma Z) = \mu + \sigma E(Z) = \mu + 0 = \mu$
$Var(Y) = Var(\mu + \sigma Z) = \sigma^2 Var(Z) = \sigma^2$
so $Y \sim N(\mu, \sigma^2)$, as required.
edit: added the explanation below
A slightly more formal way to show this is through the distribution function. Let $X, Y, Z$ be as above, and let $a$ be a real number. Then
$$
P(Y \leq a) = P(\mu + \sigma Z \leq a) = P(Z \leq \frac{a - \mu}{\sigma}) =
$$
$$
= \int_{-\infty}^{\frac{a - \mu}{\sigma}} \frac{1}{\sqrt{2\pi}} \mathrm{exp}(-\frac{z^2}{2})dz
$$
Similarly, we have
$$
P(X \leq a) = \int_{-\infty}^a \frac{1}{\sigma \sqrt{2\pi}} \mathrm{exp}(-\frac{1}{2} (\frac{x - \mu}{\sigma})^2)dx
$$
In this final integral, simply apply the substitution $z = \frac{x - \mu}{\sigma}$ to see that they are the same. In other words, $P(Y \leq a) = P(X \leq a)$ for all real numbers $a$.
