Let $V$ be a complex vector space (not necessarily finite dimensional). Let $V[[t]]\cong V\hat{\otimes}_{\mathbb C}\mathbb C[[t]]$ be the completion of $V\otimes_{\mathbb C}\mathbb C[[t]]$ with respect to the $(t)$-adic topology on $\mathbb C[[t]]$. Let $K:=\mathbb C((t))$ be the field of formal Laurent series. I would like to know what the completion of the tensor product $V[[t]]\otimes_{\mathbb C[[t]]}K$ with respect to the $(t)$-adic topology on $V[[t]]$ and the topology $\{t^i\mathbb C[[t]]\}$ on $K$ is. I come up with the result zero but I do not believe that zero is what should come out. So maybe I have made a mistake.
Here is my computation: $$V[[t]]\hat{\otimes}_{\mathbb C[[t]]}K=\lim_{i, j}(V[[t]]/(t^i)\otimes_{\mathbb C[[h]]}K/t^j\mathbb C[[t]])\\ =\lim_{i, j}\frac{V[[t]]\otimes_{\mathbb C[[t]]} K}{t^iV[[t]]\otimes_{\mathbb C[[t]]}K+ V[[t]]\otimes_{\mathbb C[[t]]}t^j\mathbb C[[t]]}\\ =\lim_{i}\frac{V[[t]]\otimes_{\mathbb C[[t]]}K}{t^iV[[t]]\otimes_{\mathbb C[[t]]}K}\\ =\lim_i\frac{V[[t]]}{(t^i)}\otimes_{\mathbb C[[t]]} K\\ =0.$$
You are correct that with your definition you get $0$. If you see people talking about getting a completion of $V((t))$ they are presumably using a different topology on the tensor product from yours, probably the one generated by neighborhoods of $0$ of the form $t^iV[[t]]\otimes t^j\mathbb{C}[[t]]$.
To give some intuition for why this should be sensible, your topology corresponds to saying that $x\otimes y$ is small if either $x$ or $y$ is small, whereas that topology would correspond to saying $x\otimes y$ is small if both $x$ and $y$ are small. But if the potential size of $y$ is unbounded (as is the case for $y\in\mathbb{C}((t))$ when $y$ could be $t^{-n}$ for arbitrarily large $n$), you shouldn't expect $x\otimes y$ to be small just because $y$ is small. Saying that $x\otimes y$ is small when $x$ is small just ends up meaning that everything is arbitrarily small, and so becomes $0$ in the completion.