The usual regression model conditions on the $x$'s. They stay the same.
$Y_i$ (not the observed $y_i$ but the random variable that $y_i$ is a realization of) has a distribution that depends on their specific $x_i$. If you and I were both to independently draw a sample with those same $x$'s we'd get different observations, and we'd get different $\hat{\beta}$ values. We'd then each have one draw from the sampling distribution of $\hat{\beta}$, some estimated coefficient $b$. We could call your realized coefficient $b^{[1]}$, and mine $b^{[2]}$ and so forth for any number of them. They're all draws from the conditional distribution (conditional on the x's) of $\hat{\beta}$.
The distribution of $\hat{\beta}$ follows from the joint conditional distribution of the $Y_i$ (the joint distribution conditional on the $x$s).
In the usual model, the vector $\mathbf{Y} = (Y_1,Y_2,...,Y_n)^\top\sim N_n(\mu,\Sigma)$ where $\mu=(\alpha+\beta x_1,\alpha+\beta x_2,...,\alpha+\beta x_n)^\top$ and $\Sigma = \sigma^2 I_n$.
The usual least squares estimator of $(\alpha,\beta)^\top$ can be written down and hence under the model assumptions, the joint distribution of $(\hat{\alpha},\hat{\beta})^\top$ and the marginal distributions of either obtained. Indeed because $(\hat{\alpha},\hat{\beta})^\top$ is linear in the Y's, they're bivariate normal and so the marginal distributions are also normal. It's a few lines of algebra to show that $E(\hat{\beta})=\beta$ and $\text{Var}(\hat{\beta})$ is as you give it.
When we carry out our exercise of drawing samples with the same x's, your estimate $b^{[1]}$, and my $b^{[2]}$ are just two draws from that normal distribution.
