The 4-dimensional spin group $Spin(4)=SU_{+}(2) \times SU_{-}(2)$, denote a typical element as $(A_+,A_-)$. We have for the 4-dimensional Euclidean space $V=\mathbb{R}^4 \simeq \mathbb{H} $ we can have the following representation $$f: Spin(4) \mapsto GL(V)$$ given by $$f(A_+,A_-)(Q)=A_-Q(A_+)^{-1}$$ Where $Q$ is a typical element of $V$ identified with $\mathbb{H}$ in the matrix form , i.e. via the following identification $$t1+xi+yj+zk \leftrightarrow \begin{bmatrix} t+xi & y+zi \\ -y+zi & t-xi \\ \end{bmatrix} $$ By calculation we can see that $f$ actually gives $f: Spin(4) \mapsto SO(V)=SO(4)$
We can also regard $Spin(4)$ as subspace of $4\times4$ matrices $\begin{bmatrix} A_+ & 0 \\ 0 & A_- \\ \end{bmatrix} $
With this, consider two copies of $\mathbb{C}^2$, i.e. $W_+$ and $W_-$ with standard hermitian metric, we have $Spin(4)$ acts on them by $$f_+(\begin{bmatrix} A_+ & 0 \\ 0 & A_- \\ \end{bmatrix})(w_+)=A_+w_+$$ $$f_+(\begin{bmatrix} A_+ & 0 \\ 0 & A_- \\ \end{bmatrix})(w_-)=A_-w_-$$
Then we can have an isomorphism of representation spaces $$V\otimes \mathbb{C} \simeq Hom(W_+,W_-)$$
Why is this true? Psychologically I understand this is simply the $\mathbb{R}^4 \otimes \mathbb{C} \simeq \mathbb{C}^4 \simeq \mathbb{C}^2 \otimes (\mathbb{C}^2)^*$, but how to establish the isomorphism in this concrete case and make it compactible with the representation given above?
By definition, $\hom(W_+,W_-)$ consists of linear maps $W_+\to W_-$ with the action
$$ (A_+,A_-)T:=A_-\circ T\circ A_+^{-1} $$
The carrier spaces (underlying sets) of $W_+$ and $W_-$ are both simply $\mathbb{C}^2$, so the linear maps $T$ may be represented by (arbitrary) $2\times 2$ complex matrices, and the $A_\pm$ are certain $2\times2$ matrices too, and composition becomes matrix multiplication. That is, $(A_+,A_-)T=A_-TA_+^{-1}$ describes the action of $\mathrm{Spin}(4):=\mathrm{SU}(2)\times\mathrm{SU}(2)$ on $\hom(W_+,W_-):= M_2(\mathbb{C})$.
$V$ is the image of the algebra embedding $\mathbb{H}\to M_2(\mathbb{C})$, with $\mathrm{SU}(2)$ the image of $S^3=\mathrm{SU}(2)$. (For comparison, $\mathrm{SO}(2)$ is the image of $\mathrm{U}(1)$ under the algebra embedding $\mathbb{C}\to M_2(\mathbb{R})$.) The action is still conjugation coming from $\mathrm{Spin}(4)$, of course, but $V$ is a real subspace of $M_2(\mathbb{C})$. However, notice that $M_2(\mathbb{C})=V\oplus iV$ is an internal direct sum of real subspaces, so we may identify $V\otimes\mathbb{C}$ with all of $M_2(\mathbb{C})$, and again the action remains conjugation!