This question relates to the construction described in this paper (the general case starts on page 8). I apologize for the third-party link, but I figured this is neater than copying everything here.
The author claims (page 9, second to last paragraph) that any continuous map $\beta X_0\to [0,1]$ is of the form $\hat{f}_0$ for some continuos function $f:X\to [0,1]$. I can't see why this is so, or more specifically, why the continuity of $f$ follows. It is clear that any continuous map $\beta X_0\to [0,1]$ induces a map $X_0\to [0,1]$ by precomposing with $\iota$, which in turn gives a map $X\to [0,1]$ by pre-composing with $\iota_0^{-1}$, but I fail to see why this last map must be continuous.
Alternatively, any continuous function $\beta X_0\to [0,1]$ induces a map $\beta X\to [0,1]$ by precomposing with the projection to the quotient $\pi$, which then gives a map $X_0\to [0,1]$ but the universal property of $\beta X_0$ gives continuity with respect to $X_0$, which is trivial since the space is discrete.
He just states this follows "by the comments preceding Theorem 2.1", by said comments refer to the universal property of the Stone-Čech compactification, which is what we're trying to prove.
Yeah, that argument is total nonsense. It makes no use of the definition of the equivalence relation $\sim$, so if it worked it would also work taking $\sim$ to be just equality, in which case $\beta X$ would be the same as $\beta X_0$ and if $\bar{\iota}$ were continuous that would mean $X$ is discrete.
Here's a different argument you could use. Suppose $F$ is an ultrafilter on $X$ converging to a point $x\in X$; to check continuity of $\bar{\iota}$, it suffices to check that $\bar{\iota}_*F$ converges to $\bar{\iota}(x)$. Considering $F$ as an ultrafilter on $X_0$, note that $\iota_*F$ converges to $F$, considering $F$ as a point of $\beta X_0$. Since $\pi$ is continuous, it follows that $\bar{\iota}_*F=\pi_*\iota_*F$ converges to $\pi(F)$, so we just need to show that $\pi(F)=\bar{\iota}(x)$. Since $\iota(x)$ is the principal ultrafilter $F_x$ at $x$, we just need to show that $F_x\sim F$. But both $F$ and $F_x$ converge to $x$ in $X$, so if $Y$ is compact Hausdorff and $f:X\to Y$ is continuous then $\hat{f}_0(F_x)=\hat{f}_0(F)=f(x)$. Thus $F_x\sim F$.