How to choose the correct volumes when preparing a dilution series?
To your questions:


  I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$.


It's always problematic expressing dilutions as ratios, because it's ambiguous as to what the numbers are referring to.  Here, the "$1$" is referring to the volume of the solution to be diluted, and the "$2$" is referring to the final volume after dilution.  Thus, the $50~\mu\mathrm L$ ("$1\times$") is being diluted to $100~\mu\mathrm L$ ("$2\times$"), and in order to do that you need to add $100~\mu\mathrm L-50~\mu\mathrm L=50~\mu\mathrm L$ of the RO $\ce{H2O}$.  This could just as validly be expressed as a $1\!:\!1$ dilution; thus the ambiguity.


  I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.


Per the same $\mathit{Initial~Volume : Final~Volume}$ ratio convention, a $1\!:\!100$ dilution to a final volume of $3~\mathrm{mL}=3000~\mu\mathrm L$ will require starting with $3000\div 100 = 30~\mu\mathrm L$ of the intermediate solution.



More generally:

I prefer to express dilutions using an "n-fold" notation, e.g.:


Dilute $50~\mu\mathrm L$ of dye solution $2$-fold with RO $\ce{H2O}$
Dilute $30~\mu\mathrm L$ of the solution from step ($1$) $100$-fold with RO $\ce{H2O}$


Though there is still room for ambiguity in this phrasing, to me diluting $n$-fold indicates much more clearly that the final volume $V_f$ should be $n\times$ the initial volume $V_i$, and thus that the volume of diluent needed is:

$$
V_d = V_f-V_i = nV_i-V_i = V_i\left(n-1\right)
$$