The set of points that satisfy the equations $f_1(x)=f_2(x)=f_3(x)=0$ can be described as the image of the map
$$
\mathbb{R}^2\rightarrow\mathbb{R}^4,
$$
where
$$
(s,t)\mapsto(s^3,s^2t,st^2,t^3).
$$
Plugging these points into the Jacobian matrix that you wrote above gives
$$
\begin{bmatrix}
st^2&-2s^2t&s^3&0\\
0&t^3&-2st^2&s^2t\\
t^3&-st^2&-s^2t&s^3
\end{bmatrix}
$$
Now, observe that $t$ times row $1$ minus $s$ times row $3$ is
$$
\begin{bmatrix}st^3&-2s^2t^2&s^3t&0\end{bmatrix}-\begin{bmatrix}st^3&-s^2t^2&-s^3t&s^4\end{bmatrix}=\begin{bmatrix}0&-s^2t^2&2s^3t&-s^4\end{bmatrix},
$$
which is $-\frac{s^2}{t}$ times row $2$. So, the rank of this image is $2$. At all (most?) other points, the rank is $3$ (this is what is meant by generically).
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.
Best Answer
For every $x$, $rank(DF_x)\leq 2$. Now $rank(dF_x)\leq 1$ iff there is $\lambda$ s.t.
$(*)$ $x_2=2\lambda x_1,x_1=2\lambda x_2,x_4=-2\lambda x_3,x_3=-2\lambda x_4$ (since $x\in M$, $x\not=0$).
Consider the cases when $\lambda=\pm 1/2$ and $\lambda\not=\pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.