I have a serious misunderstanding of tensor products. I think If I can understand these two examples, then most of the difficulties will be eliminated.
$0^{\rm th}$ Question (Edited): Let $k$ be a field, and let $A$ and $B$ and $C$ be finite-dimensional $k$-algebras. To prove that $A \otimes B \cong C$, is it sufficient to give a multiplication-preserving bijection between the elements of basis as a $k$-vector-space, or as a $B$-module?
First and main question: Let $K/k$ be a Galois extension of fields, with the Galois group $G$. Then $K\otimes_k K \cong K^G$, $x\otimes y \mapsto \Big(\sigma(x)y\Big)_{\sigma \in G}$.
Let's call that map by $\varphi$, then it is easy to see that
$$\varphi((x\otimes 1)(x'\otimes 1))=\varphi(xx'\otimes 1)=\Big(\sigma(xx')\Big)_{\sigma \in G}=\Big(\sigma(x)\Big)_{\sigma \in G}\Big(\sigma(x')\Big)_{\sigma \in G}=\varphi(x\otimes 1)\varphi(x'\otimes 1)$$
Why didn't we defined the map as follows? $K\otimes_k K \cong K^G$, $x\otimes y \mapsto \Big(xy\Big)_{\sigma \in G}$
Let's call that map by $\varphi_{\rm id}$, then again we have
$$\varphi_{\rm id}((x\otimes 1)(x'\otimes 1))=\varphi_{\rm id}(xx'\otimes 1)=\Big(xx'\Big)_{\sigma \in G}=\Big(x\Big)_{\sigma \in G}\Big(x'\Big)_{\sigma \in G}=\varphi_{\rm id}(x\otimes 1)\varphi_{\rm id}(x'\otimes 1)$$
What is wrong with them in this situation?
Or why didn't we defined the map as follows? Fix $\tau \in G$, then $K\otimes_k K \cong K^G$, $x\otimes y \mapsto \Big(\tau(x)y\Big)_{\sigma \in G}$
Let's call that map by $\varphi_{\tau}$, then again we have
$$\varphi_{\tau}((x\otimes 1)(x'\otimes 1))=\varphi_{\tau}(xx'\otimes 1)=\Big(\tau(xx')\Big)_{\sigma \in G}=\Big(\tau(x)\Big)_{\sigma \in G}\Big(\tau(x')\Big)_{\sigma \in G}=\varphi_{\tau}(x\otimes 1)\varphi_{\tau}(x'\otimes 1)$$
What is wrong with them in this situation?
Second question: I asked another question before here: $M_n(k)\otimes_kB\cong M_n(B)$, where $k$ is a field, and $B$ is a $k$-algebra.
Let $k$ be a field, and $A$ and $B$ be $k$-algebras. Then $M_n(A)\otimes_kB\cong M_n(A\otimes B$). (I am not sure if my solution is true or not)
$M_n(A)\otimes_kB$ is a free $B$-module, with the basis $\{(E_{ij}\otimes1_B) \mid 1 \leq i, j \leq n \}$. Now define the map $\varphi$ on the elements of basis as follows:
\begin{gather*} \varphi:M_n(A) \otimes_k B \longrightarrow M_n(A\otimes B)\\ (E_{ij}\otimes1_B) \mapsto E_{ij} \end{gather*}
Then we have
$$\varphi((E_{ij}\otimes1_B)(E_{kl}\otimes1_B))=\varphi(E_{ij}E_{kl}\otimes 1_B)=\varphi(\delta_k^jE_{il}\otimes 1_B)=\delta_k^jE_{il}=E_{ij}E_{kl}=\varphi(E_{ij}\otimes1_B)\varphi(E_{kl}\otimes1_B)$$
Is it true?
The map $K\otimes_kK\to K^G$ defined by $x\otimes y\mapsto(xy)_{\sigma\in G}$ is not an isomorphism (as long as $|K:k|\ge2$). Every element $x\otimes1-1\otimes x$ lies in its kernel, and that element is nonzero if $x\in K\setminus k$.