Wikipedia gives the following explicit formula for the functional calculus of $2\times2$ matrices:
$$
f(A) = \frac{f(\lambda_+) + f(\lambda_-)}{2} I + \frac{\mathrm{tr}(A)/2 - \mathrm{adj}(A)}{\sqrt{\Delta(A)}} \frac{f(\lambda_+) - f(\lambda_-)}{2},
$$
where I am guessing that $\mathrm{tr}(A)/2 - \mathrm{adj}(A)$ actually means $(\mathrm{tr}(A)/2) I - \mathrm{adj}(A)$ and $\mathrm{adj}(A)$ stands for the adjugate of $A$ and where $\Delta(A)=(\mathrm{tr}(A)/2)^2 - \mathrm{det}(A)$ is the discriminant of the characteristic polynomial $\chi_A(t)=\mathrm{det}(tI-A)=t^2-\mathrm{tr}(A)t+\mathrm{det}(A)$ and $\lambda_\pm = \mathrm{tr}(A)/2 \pm \sqrt{\Delta(A)}$ are its roots. Unfortunately, the Wikipedia article is rather *cryptic* in this place without listing any assumptions whatsoever, and I have not been able to locate the formula in either Higham's book or in Bhatia's.

Is (my interpretation of) this formula really correct for any $A\in M_2(\mathbb{C})$ with $\Delta(A)\neq 0$ (suffices to assume $f\in\mathbb{C}[t]$ at the moment)? If not, then in what generality and where can I find a rigorous derivation of it?

any$2\times2$ matrix, and the answer to that is trivially no: it doesn't work for $f$ the identity function and $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. The functional calculus can only be expected to work for normal matrices. $\endgroup$