I am trying to prove that the evaluation homomorphism:
$ev_z:\mathbb{R}[X]\to \mathbb{C}, f\mapsto f(z)$ where $z=a+bi$ is surjective.
To start with, I don't really understand how the map is obtained. Do have a polynomial and we just change $x$ to $z=a+bi$? But how do we determine $a$ and $b$ for $z$?
EDIT: My original answer was completely wrong, as I misread the question. My mistake was helpfully pointed out in the comments, and I have corrected it.
You are right that the map is obtained by substituting $z$ for $x$ in the expression of the polynomial. As for surjectivity, I'd note that the result is only true if $b \neq 0$, for if $z \in \mathbb{R}$, then $f(z) \in \mathbb{R}$ for any real polynomial. So we assume $b \neq 0$, and proceed as follows.
We have $ev_z(1) = 1$ and $ev_z(X) = z = a + bi$. The latter gives us that
$$ ev_z(\frac{X-a}{b}) = i $$
It is pretty clear then that for any real $u, v$, we have \begin{align} u + vi &= ev_z(u) + v\cdot ev_z(\frac{X - a}{b}) \\ &= ev_z(u + v\frac{X-a}{b}) \\ &= ev_z\big(\frac{v}{b}X + (u-\frac{av}{b})\big) \end{align}
Since every complex number is of the form $u + vi$ for some $u, v$, we have that $ev_z$ is surjective.
tl;dr for any $u + vi \in\mathbb C$, we have $f(X) = \frac{v}{b}X + (u-\frac{av}{b}) \in \mathbb{R}[X]$, and it is trivial to check that $$ f(a + bi) = \frac{v}{b}(a + bi) + (u - \frac{av}{b}) = 0 $$