Let $R$ be a ring (commutative with unit).Let $f_1,...,f_n\in R$ elements that generate the unit ideal. The map $R\to R_{f_1}\times ...\times R_{f_n}$ is faithfully flat, since this is just a Zariski cover. Now, my question is

Is the map $R[[t]]\to R_{f_1}[[t]]\times ...\times R_{f_n}[[t]]$ faithfully flat?

Note that in general $R_f [[t]]\ne R[[t]]_f$, so it is *not* a special case of the previous map.

I general, I am quite confused about how to work with rings like $R[[t]]$. Objects such as $R_f[[t]]$ seem much more natural then, say, $R[[t]]_f$. Yet abstractly, $R[[t]]_f$ is just a localization, where $R_f[[t]]$ is something more complicated. I guess $R_f[[t]]$ is a "completed" version of $R[[t]]_f$, but what is the formal description that allows one to derive properties of it?

**Edit**: As said in the comments, every maximal ideal of $A=R[[t]]$ is of the form $\mathfrak{m}+(t)$ and hence its extension in $B=R_{f_1}[[t]]\times ...\times R_{f_n}[[t]]$ is not the unit ideal (looking at the factor where $f_i \not\in \mathfrak{m}$). So the question is whether $A\to B$ is flat or just whether $R[[t]]\to R_f[[t]]$ is flat, since this is trivially equivalent.