Faithfully Flat Ring Homomorphism of Power Series

The usual way to show that a map of this kind is faithfully flat is to use the Artin--Rees Lemma and its corollaries, which show that adic completions of Noetherian rings are flat, and faithfully flat under appropriate hypotheses. For example, the completion of a Noetherian ring at any ideal contained in its Jacobson radical is faithfully flat.

In your particular case, you can check that the target of your inclusion $f$ is the $x$-adic completion of the source, and so the map is flat.

However, $x$ is not in the Jacobson radical of $R[[x]][y]$ (although it is in the Jacobson radical of $R[[x]]$, and also in the Jacobson radical of $R[y][[x]]$).

E.g. let $\mathfrak m_R$ be the maximal ideal of $R$, and consider the ideal $\mathfrak m := (\mathfrak m_R, xy -1)$. The quotient $R[[x]][y]/\mathfrak m$ is equal to $(R/\mathfrak m_R)[[x]][1/x]$, which is the field of Laurent series in $x$ over the residue field $R/\mathfrak m_R$. Thus $\mathfrak m$ is maximal, but does not contain $x$.

By Artin--Rees, the tensor product of $R[y][[x]]$ with $R[[x]][y]/\mathfrak m$ over $R[[x]][y]$ is equal to the $x$-adic completion of $R[[x]][y]/\mathfrak m$, which vanishes. Thus $R[y][[x]]$ is not faithfully flat over $R[[x]][y]$.

In fact, we could have just looked at the ideal $I = (xy - 1)$; i.e. the tensor product $R[y][[x]]\otimes_{R[[x]][y]} (R[[x]][y]/I)$ already vanishes, because $1 - xy$ is a unit in $R[y][[x]]$. The reason I introduced $\mathfrak m$ at all is just to show explicitly that $x$ is not in the Jacobson radical of $R[[x]][y]$.

The basic intuition is that in $R[[x]][y]$ you are allowed to specialize $y$ to be $1/x$. But in $R[y][[x]]$ you have elements of the form $ 1 + x y + x^2 y^2 + \cdots + x^n y^n + \cdots$ (this is precisely the inverse of $1 - xy$) and you cannot substitute $y = 1/x$ into such an element in a meaningful way.