But you told the OP to consider this isomorphism $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$

$End_\Bbb{C}(\Bbb{C}^2)=M_2(\Bbb{C})$, the $2\times 2$ complex matrices.

I tried to collect what is unclear to me in all the questions I asked above. The OP did not specify at the begining the tensor product is on which field and then he said on one of his comment ... ok it the real field.

I am aware that you are using this idea: $\mathbb{H} \subset M_{2}(\mathbb{C})$ is the set of matrices of the form: $$ \begin{pmatrix} z & - \bar{\omega} \\ \omega & \bar{z} \end{pmatrix} \quad \quad z, \omega \in \mathbb{C}$$

And then for the isomorphism: if $z = a + ib, \omega = c + id,$we can rewrite $\mathbb{H}$ as: $$ \begin{pmatrix} z & - \bar{\omega} \\ \omega & \bar{z} \end{pmatrix} = \begin{pmatrix} a+ib & -c + id \\ c+id & a-ib \end{pmatrix} = a\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} + c\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} = aI + bJ + c K + d L$$

Where, $$ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, J = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, K =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, L = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}.$$

Yes but no, for $\Bbb{H}$ then $a,b,c,d$ are real numbers, in $\Bbb{H\otimes_RC}$ they are complex numbers, so $\{aI+bJ+cK+dL,a,b,c,d\in \Bbb{C}\}$ is the whole of $M_2(\Bbb{C})$. That what we wanted to prove: the isomorphism between $\Bbb{H\otimes_RC}$ and $M_2(\Bbb{C})$.

I also, got confused from the other question I read, and sorry if that is a provocating question, how to prove that the map you gave for the OP is surjective?

So you are saying that the OP question reduces to proving the isomorphism you mentioned in your last comment?

Yes, that's why my answer takes only 3 lines. $\frac{a}2 I-i\frac{a}2 J = \pmatrix{a&0\\0&0}$, following the same idea it is clear that it is surjective.

You didn't mention the ring (complex algebra) structure on $\Bbb{H\otimes_R C}$ I am quite sure this is the least obvious part.

When you say "following the same idea" you mean which idea?

If you could explain this ring (complex algebra) structure on $\mathbb H \otimes_R \mathbb C$ , I will be beyond grateful to that.

I am really sorry but posting an incomplete question and asking people to complete its ingredients (which the OP did) is really confusing to me

Is this the ring structure you are referring to? what about my comment before it?

I did not get your idea in this statement "The $i$ of $i\otimes b$, the $i$ hidden in $a=A+iB$, and the $i$ in $\pmatrix{i&0\\0&-i}$ are not the same, that's part of the game to understand what we identity the latter two." How the $i's$ are different ? what do you mean by " we identity the latter two."

Also, I see that you have chosen $\Bbb{H}$ to be a 2-dimensional (left) vector space and you did not take it as a 2-dimensional (right) vector space ..... it seems to me that left and right is only by your choice and there is nothing in the problem oblige you to that choice.

should not the isomorphism first be with just $\times$ and then the UMP of tensor product induces the map with $\oplus$?

$End_\Bbb{C}(\Bbb{H})$ means that we assume some complex vector space structure on $\Bbb{H}$. And you can check that the natural ones (ie. compatible with the canonical real vector space structure) all give $\Bbb{H}\cong \Bbb{C}^2$, independently of left right, or the swapping of $i,j$, or the basis.