Recall that if a field $k$ has a primitive $n$-root of unity $\omega$, then the cyclic $k$-alegbras of degree $n$ (ie of dimension $n^2$) have the following familiar presentation : they are generated by two elements $u$ and $v$ satisfying $u^4 = a\in k^*$, $v^4 = b\in k^*$ and $vu=\omega uv$. This algebra will be denoted by $(a,b)_{n,\omega}$ (the choice of primitive root $\omega$ is mostly benign).
For $n=4$, It's not very hard to prove that $(a,b)_{4,\omega}\otimes_k (a,b)_{4,\omega}\simeq M_4\left( (a,b^2)_{4,\omega}\right)\simeq M_8\left( (a,b)_2\right)$, where $(a,b)_2$ is the usual quaternion algebra defined by $a$ and $b$ over $k$ (in this case we can omit the root of unity since it can only be $-1$).
The isomorphism $\Phi: (a,b^2)_{4,\omega}\to M_2\left( (a,b)_2\right)$ is given for instance by $u\mapsto \begin{pmatrix}0 & i \\ 1 & 0 \end{pmatrix}$ and $v\mapsto \begin{pmatrix}j & 0 \\ 0 & \omega j \end{pmatrix}$ where $i$ and $j$ are the usual quaternion generators (satisfying $i^2 = a$, $j^2=b$ and $ij = -ji$).
In particular, if $(a,b)_2$ is split, then $(a,b)_{4,\omega}$ has period $2$, so by a classical result of Albert it's a biquaternion algebra : $(a,b)_{4,\omega}\simeq (x_1,y_1)_2\otimes_k (x_2,y_2)_2$.
Given an explicit reason why $(a,b)_2$ is split, namely given explicit elements $\lambda,\mu\in k^*$ such that $b = \lambda^2-a\mu^2$, can we find an explicit formula for some $x_i$ and $y_i$ ?
Reverse-engineering the general proof of Albert's theorem leads to pretty atrocious calculations, but I feel that given this explicit presentation there should be a relatively simple answer.
In the special case $\mu=0$, we get $(a,b)_{4,\omega}\simeq (a,\lambda)_2\otimes_k (1,1)_2$, replacing $b$ with $\lambda$ in the isomorphism $\Phi$ above (since $A\otimes_k (1,1)_2 \simeq M_2(A)$ for any algebra $A$). But this is somewhat ad hoc, and I don't see how to generalize it.
Through some calculations I managed to find an answer : if $b = \lambda^2 - a \mu^2$, then $$(a,b)_{4,\omega} \simeq (a,\lambda)_2 \otimes_k (1,1)_2 \simeq M_2\left( (a,\lambda)_2 \right) $$ with an explicit isomorphism given by $$\begin{array}{rrcl} \Phi: & (a,2\omega b\lambda)_2 \otimes_k (1,1)_2 & \to & (a,b)_{4,\omega} \\ & i_1\otimes 1 & \mapsto & u^2 \\ & j_1 \otimes 1 & \mapsto & \omega (\lambda + \mu u^2)v + v^3 & \\ & 1 \otimes i_2 & \mapsto & \frac{1}{2}u(1+ \frac{1}{b}v^2) + \frac{1}{2a}u^3(1 - \frac{1}{b}v^2)& \\ & 1 \otimes j_2 & \mapsto & \frac{1}{b}v^2 & \\ \end{array}$$ where $i_1$ and $j_1$ (resp. $i_2$ and $j_2$) are the canonical generators of $(a,\lambda)_2$ (resp. $(1,1)_2$).
This concludes since $(a,2\omega b\lambda)_2 \simeq (a,\lambda)_2$ (but writing directly an isomorphism $(a,\lambda)_2 \otimes_k (1,1)_2 \to (a,b)_{4,\omega}$ gives a uselessly complicated formula).