I haven't done linear algebra for a while, so, please bear with me. Let $d\in\mathbb N$ and $U$ be a $(d-1)$-dimensional subspace of $\mathbb R^d$. It's easy to show that there is some $\eta$ in the algebraic dual space of $\mathbb R^d$ with $$U=\ker\eta\;.\tag1$$ Now, let $P:\mathbb R^d\to\mathbb R$ be a polynomial of degree $d$ with $$\left.P\right|_U=0\;.\tag2$$
I want to show that there is a polynomial $Q:\mathbb R^d\to\mathbb R$ of degree $d-1$ with $$P=\eta Q\;.\tag3$$
I struggle with the question why we can assume that $$\eta(x)=x_d\;\;\;\text{for all }x\in\mathbb R^d\tag3$$ and $$U=\mathbb R^{d-1}\times\left\{0\right\}\tag4\;.$$ Actually, it's intuitively clear to me, but I don't remind how we need to argue rigorously. Sure, we somehow need to find a function $T:\mathbb R^d\to\mathbb R^d$ (which rotates and translates $U$) such that $$T(U)=\mathbb R^{d-1}\times\left\{0\right\}\tag5\;.$$ How can we do that and why is the assumption of $(3)$ and $(4)$ legitimate?
Some basic facts about polynomials and linear transformations on $\mathbb R^d:$
Lemma 1: If $P$ is a polynomial on $\mathbb R^d$ of degree $n$ and $T:\mathbb R^d \to \mathbb R^d$ is a nonsingular linear transformation, then $P\circ T$ is a polynomial on $\mathbb R^d$ of degree $n.$
Lemma 2: If $P(x_1,\dots,x_d)$ is a polynomial on $\mathbb R^d$ of degree $n>0$ and $P=0$ on $\mathbb R^{d-1}\times \{0\},$ then there exists a polynomial $Q$ of degree $n-1$ such that $P(x) = x_dQ(x)$ for all $x\in \mathbb R^d.$
Lemma 3: If $U$ is a $(d-1)$ dimensional subspace of $\mathbb R^d,$ and $\eta_1, \eta_2:\mathbb R^d \to \mathbb R$ are linear maps with $\ker\eta_1= \ker \eta_2=U,$ then there is a nonzero constant $c\in \mathbb R$ such that $\eta_2=c \eta_1$ on $\mathbb R^d.$
Back to the problem, where we have $P=0$ on $U = \ker \eta,$ with $P$ a polynomial of degree $n>0.$ (I'm not sure why you have $n=d$ in the problem; the result is true for arbitrary $n.$)
Choose a basis $u_1,\dots,u_{d-1}$ for $U,$ and extend it to a basis $u_1,\dots,u_{d-1},u_d$ of $\mathbb R^d.$ Let $e_1,\dots ,e_d$ be the standard basis for $\mathbb R^d.$ Then there is a unique nonsingular linear $T:\mathbb R^d \to \mathbb R^d$ such that $T(e_k) = v_k$ for all $k.$ We have $T$ mapping $\mathbb R^{d-1}\times \{0\}$ isomorphically onto $U,$ with $T^{-1}$ "doing the same" coming back.
By Lemma 1, $P\circ T$ is a polynomial of degree $n>0$ that vanishes on $\mathbb R^{d-1}\times \{0\}.$ By Lemma 2, $P\circ T(x) = x_dQ(x)$ for some polynomial $Q$ of degree $n-1.$ Let $\Pi(x) = x_d, x\in \mathbb R^d.$ With this notation we have $P\circ T = \Pi\cdot Q$ on $\mathbb R^d.$ Therefore $P = (\Pi\circ T^{-1})\cdot (Q \circ T^{-1}).$ Note that $Q \circ T^{-1}$ is a polynomial of degee $n-1$ by Lemma 2.
Because $T^{-1}$ maps $U$ to $\mathbb R^{d-1}\times \{0\},$ and $\Pi =0$ on this last set, we have $\Pi\circ T^{-1} = c\eta$ by Lemma 3. Thus $P = (c\eta)\cdot (Q \circ T^{-1}) = \eta\cdot (cQ \circ T^{-1}).$ This shows $P$ has the desired form and we're done.