I am not going to do a rigorous proof. But what follows is a demonstration that there is a mapping between any hyperduel and any degree 2 polynomial such that multiplication results the same. This can be extended to other operations.
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Basic definitions:
Hyperduels:
$$a=a_{0}+a_{1}\epsilon_{1}+a_{2}\epsilon_{2}+a_{3}\epsilon_{1}\epsilon_{2}$$
$$b=b_{0}+b_{1}\epsilon_{1}+b_{2}\epsilon_{2}+b_{3}\epsilon_{1}\epsilon_{2}$$
$$ab=a_{0}b_{0}+\left(a_{1}b_{0}+a_{0}b_{1}\right)\epsilon_{1}+\left(a_{2}b_{0}+a_{0}b_{2}\right)\epsilon_{2}+\left(a_{0}b_{3}+a_{1}b_{2}+a_{2}b_{1}+a_{3}b_{0}\right)\epsilon_{1}\epsilon_{2}$$
Polynomials of degree 2:
$$w=w_{0}+w_{1}x^{1}+w_{2}x^{2}$$
$$u=u_{0}+u_{1}x^{1}+u_{2}x^{2}$$
$$uw=w_{0}u_{0}+\left(w_{0}u_{1}+w_{1}u_{0}\right)x^{1}+(w_{0}u_{2}+w_{1}u_{1}+w_{2}u_{0})x^{2}$$
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The first thing to note is:
The only use of hyperduels uses in differentiation:
$a_{1}=a_{2}$, and $b_{1}=b_{2}$ since they are both equal to first derivative of something (the initial perturbation).
(If you do not hold this then hyperduels and polynomials of degree 2 are not the same thing. Since one has a extra free variable)
Thus actually:
$$a=a_{0}+a_{1}\epsilon_{1}+a_{1}\epsilon_{2}+a_{3}\epsilon_{1}\epsilon_{2}$$
Thus:
$$ab=a_{0}b_{0}+\left(a_{1}b_{0}+a_{0}b_{1}\right)\epsilon_{1}+\left(a_{1}b_{0}+a_{0}b_{1}\right)\epsilon_{2}+\left(a_{0}b_{3}+2a_{1}b_{1}+a_{3}b_{0}\right)\epsilon_{1}\epsilon_{2}$$
Now:
We declare a mapping between Polynomials of Degree 2 and HyerDuels
$$h\mapsto h'$$
$$h_{0}+h_{1}x+h_{2}x^{2}\mapsto h_{0}+h_{1}\epsilon_{1}+h_{1}\epsilon_{2}+2(h_{2}-h_{1})\epsilon_{1}\epsilon_{2}$$
Under this:
$$w^{\prime}=w_{0}+w_{1}\epsilon_{1}+w_{1}\epsilon_{2}+2(w_{2}-w_{1})\epsilon_{1}\epsilon_{2}$$
$$u^{\prime}=u_{0}+u_{1}\epsilon_{1}+u_{1}\epsilon_{2}+2(u_{2}-u_{1})\epsilon_{1}\epsilon_{2}$$
$$w^{\prime}u^{\prime}=w_{0}u_{0}+\left(w_{0}u_{1}+w_{1}u_{0}\right)\epsilon_{1}+\left(w_{0}u_{1}+w_{1}u_{0}\right)\epsilon_{2}+\left(2\left(u_{2}w_{0}-u_{1}w_{0}\right)+2u_{1}w_{1}+2\left(w_{2}u_{0}-w_{1}u_{0}\right)\right)\epsilon_{1}\epsilon_{2}$$
$$w^{\prime}u^{\prime}=w_{0}u_{0}+\left(w_{0}u_{1}+w_{1}u_{0}\right)\epsilon_{1}+\left(w_{0}u_{1}+w_{1}u_{0}\right)\epsilon_{2}+2\left(\left(u_{2}w_{0}+u_{1}w_{1}+w_{2}u_{0}\right)-(w_{1}u_{0}+u_{1}w_{0}\right)\epsilon_{1}\epsilon_{2}$$
taking the mapping back the other way we see:
$$\left(w^{\prime}u^{\prime}\right)^{-\prime}=w_{0}u_{0}+\left(w_{0}u_{1}+w_{1}u_{0}\right)x+\left(u_{2}w_{0}+u_{1}w_{1}+w_{2}u_{0}\right)x^{2}$$
Which is the same as: $$uw=w_{0}u_{0}+\left(w_{0}u_{1}+w_{1}u_{0}\right)x^{1}+(w_{0}u_{2}+w_{1}u_{1}+w_{2}u_{0})x^{2}$$ which we saw before.
One can go through and show they have matching addition operations similarly.
And then show their multiplicative and additive identities are equal.
Until you have demonstrated they are the same thing for what ever definition of same you require.
