For $ k = 0, 1, \ldots, n-1$, consider the (horizontal) vector $v_k$ with $i$th coordinate $$ \sum \prod_{j=1, a_j \neq i}^{k} {x_{a_j}}.$$
For example, with $n = 3$, we have
$v_0 = (1, 1, 1)$,
$v_1 = (x_2 + x_3, x_3 + x_1, x_1 + x_2)$,
$v_2 = ( x_2x_3, x_3x_1, x_1x_2)$.
With $n = 4$, we have
$v_0 = (1, 1, 1, 1)$,
$v_1 = (x_2 + x_3 + x_4, x_3 + x_4 + x_1, x_4 + x_1 + x_2, x_1 + x_2 + x_3)$,
$v_2 = ( x_2x_3+x_3x_4+x_4x_2, x_3x_4+x_4x_1+x_1x_3, x_4x_1+x_1x_2+x_2x_4, x_1x_2 + x_2x_3 + x_3x_2)$,
$v_3 = (x_2x_3x_4, x_3x_4x_1, x_4x_1x_2, x_1x_2x_3)$.
Claim: $v_kA = (n-1-k) v_k$.
Proof: Expand it. A lot of the cross terms cancel out.
For example, with $v_0$, the column sum is $n-1$, so $v_0 A = (n-1) v_0$.
For example, with $v_{k-1}$, the numerators are all $\prod x_i$, and by looking at the denomninators, they cancel out to 0, so $v_{k-1} A = 0 $.
Do you see how we get $v_k A = (n-1-k)v_k$?
Corollary: The eigenvalues are $0, 1, 2, \ldots, n-1 $.
Suppose $b= a/2+c$ where $c>0$. Then we have
$$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$
This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$
has exactly two minimums with negative value, say at $x_1$ and $x_3$ where
$x_1<x_3$, so $p(x_1)<0$ and $p(x_3)<0$ i.e.
$$x_i^4-x_i^3+ax_i+{a\over 2} <0\;\;\;(*)$$
for $i=1,3$. Since $x_1,x_3$ satisfies also $p'(x_i)=0$ we have $$a=-4x_i^3+3x_i^2\;\;\;(**)$$ so we
get plugging $(**)$ in inequality $(*)$ for $i=1,3$:
$$-6x_i^4+3x_i^2<0\implies x_i^2>{1\over 2}$$
If we subtract equations $(**)$, we get
$$(x_1-x_3)\Big(4(x_1^2+x_1x_3+x_3^2)-3(x_1+x_3)\Big)=0$$
so $$4(x_1^2+x_1x_3+x_3^2)=3(x_1+x_3)$$ and from here we get
$$2(x_1^2+x_3^2)+2(x_1+x_3)^2=3(x_1+x_3)$$ which means that $$
2+2t^2<3t$$ where $t=x_1+x_3$. But this inequality does not have a
solution so we have a contradiction.
Best Answer
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either $$d_1d_3>0,\qquad d_1d_4>0,\qquad d_2d_3>0,\qquad\text{ or }\qquad d_2d_4>0.$$ Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy $$d_1+d_2+d_3+d_4=k,\qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.\tag{1}$$ For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $k\neq0$ we must have $d_i\neq0$ for some $i$. If $d_1\neq0$ or $d_2\neq0$ then $d_3=d_4=0$, and if $d_3\neq0$ or $d_4\neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to $$d_1+d_2=k,\quad d_3=d_4=0\qquad\text{ or }\qquad d_3+d_4=k,\quad d_1=d_2=0,$$ which is precisely twice the number of solutions to $a+b=k$ with $a,b\in\Bbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form $$x_1^n,\qquad x_2^n,\qquad x_3^n,\qquad x_4^n,\qquad x_1^mx_2^n,\qquad x_3^mx_4^n.$$ Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.