Yes, you are correct. $A [[t]] \simeq A \mathbin{\hat\otimes} \Bbbk [[t]]$ and $A \otimes \Bbbk [[t]]$ without $\hat{}$ consists of the formal power series in $t$ whose coefficients span a finite-dimensional subspace of $A$.
Long answer. It seems like you are looking at deformations of associative algebras, where $A$ is the $\Bbbk$-algebra to be deformed and $B$ is your base of deformation — in your example $B = \Bbbk [[t]]$. (If not, let me just illustrate the discussion for this case.)
If either $A$ or $B$ are finite-dimensional, then $A \mathbin{\hat\otimes} B = A \otimes B$, but if both $A$ and $B$ are infinite-dimensional one should work with $\hat\otimes$.
Since classical deformation functors are often defined on (commutative) local Artinian $\Bbbk$-algebras (which are finite-dimensional), it is enough to look at $A \otimes B$ when $B$ is local Artinian. (These are sometimes called infitesimal deformations.)
However, more generally a formal deformation over $B$ should be given by an associative algebra structure on $A \mathbin{\hat\otimes} B$, where $B$ is a complete local Noetherian $\Bbbk$-algebra such that modulo the maximal ideal $\mathfrak m$ of $B$ you recover the original algebra structure on $A$. Since not every complete local Noetherian $\Bbbk$-algebra is finite-dimensional, one should really write $\hat\otimes$ when $A$ is not necessarily finite-dimensional, but the extra $\hat{}$ is easy to forget.
A nice source for reading about this is Chapter 1 in:
- Markl, Martin, Deformation theory of algebras and their diagrams, CBMS Regional Conference Series in Mathematics 116. Providence, RI: American Mathematical Society (AMS); Washington, DC: Conference Board of the Mathematical Sciences (CBMS) (ISBN 978-0-8218-8979-4/pbk). ix, 129 p. (2012). ZBL1267.16029.
\hat
on\otimes
cancels out its\mathbin
ness, which you have manually to re-instate. Compare $A \hat\otimes k[[t]]$A \hat{\otimes} k[[t]]
to $A \mathbin{\hat\otimes} k[[t]]$A \mathbin{\hat\otimes} k[[t]]
. I have edited accordingly. $\endgroup$