p. 6: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf
Pretend the blue set was not given and I have to calculate it myself:
For all $x \in G, f(x) = xgH$ is faithful $\iff \color{blue}{\text{some set which I must find}} = \color{green}{\{id\}} $.
The PDF unfolds the answer: Because $ {g_2}^{-1}g_1 \in \color{blue}{\bigcap_{g \in G} gHg^{-1}} $, hence $ \color{blue}{\bigcap_{g \in G} gHg^{-1}} = \color{green}{\{id\}} $.
I tried: To determine the values of $ x $ for which $ f(x) = xgH$ is faithful, I solve $ g_1 * x = g_2 * x $. Here I let * represent the group binary operation.
$ g_1 * x = g_2 * x :\iff g_1(gH) = g_2(gH) \text{ for all } g \in G \iff g_2^{-1}g_1 \in \color{red}{gH}$.
The last $\iff$ is by dint of the result: ${ aH = bH \iff b^{-1}a \in H} $.
What did I bungle? I missed $\color{red}{\bigcap_{g \in G}}$ and $g^{-1}$?