I want to understand the meaning of the following sentence in terms of tensor product and group action, not in the essence.
$(A \otimes_{\mathbb{Z}} R)[[t]]$ has an $R$-linear action of $\phi$, where $\phi$ is the Frobenius on the ring of witt vectors $A$ over a residue field $\kappa$.
Here $A$ is the ring Witt vector vectors over a characteristic $p$ residue field $\kappa$ and $ R$ is a ring.
Answer:
Since $\phi$ is Frobenius on $A$, we have $\phi(t)=t^p, \ t \in A$.
Then the line "$(A \otimes_{\mathbb{Z}} R)[[t]]$ has an $R$-linear action of $\phi$" means $\exists \ $ linear map $\phi: A \otimes R \to A$ defined by $\phi((x,r))=\phi(x \otimes r)=\phi(x)r=x^pr$ is linear map, where $x \in A$ and $r \in R$.
Am I true?
Please explain the line in general sense, at least.
This is a very general phenomenon, there is nothing special going on here: if $A$ and $R$ are any rings, and $\phi$ is any endomorphism of $A$, then it induces an endomorphism of $A\otimes_\mathbb{Z} R$ by $$\phi(a\otimes r) = \phi(a)\otimes r$$ which is $R$-linear by construction.
This extends naturally to $(A\otimes_\mathbb{Z} R)[[t]]$ by making $\phi$ act on the coefficients of a power series, and it is still $R$-linear by construction.
In particular, this has nothing to do with a map $A\otimes_\mathbb{Z} R\to A$ (unless there is more context you did not tell about). Also, be careful that the Frobenius is not (at all) $x\mapsto x^p$ on $A$.