As a preface I don't much background in this area and I think I am dealing with older (or currently non-standard) definitions.
Definition: A polynomial $f \in \mathbb{R}[x_1, \ldots, x_n]$ is homogenous of degree $d$ if for all $\lambda \in \mathbb{R}, f(\lambda x_1, \ldots, \lambda x_n) = \lambda^d f(x_1, \ldots, x_n)$.
Definition: An ideal $I \subset \mathbb{R}[x_1, \ldots, x_n]$ is homogenous if it is generated by finitely many homogenous polynomials.
Definition: $X \subset \mathbb{P}(\mathbb{R}^n)$ is a real projective variety if it is the zero locus of a prime homogenous ideal $I \subset \mathbb{R}[x_1, \ldots, x_n]$.
Now for my question, I have a $X \subset \mathbb{P}(\mathbb{R}^n)$ which is a real projective variety. I have a polynomial map $F = (f_1, \ldots, f_k)$ where $f_i : \mathbb{R}^n \to \mathbb{R}$ are homogenous monomials of degree $d$, with $d < n$. I am interested in understanding the image $F(X)$. In particular, I care about when $F(X)$ is closed with respect to the Zariski Topology on $\mathbb{P}(\mathbb{R}^k)$.
From reading I know that if we were working over an algebraically closed field then $F$ is a closed and hence $F(X)$ is a variety, as discussed here. However, in my case we are working over $\mathbb{R}$. I cannot seem to find anything about the case when the field is not algebraically closed.
For more context, for a given $m$, I have matrices $(A_1, \ldots, A_m) \in \mathbb{P}(\mathbb{R}^{m \times M \times N})$ such that $A_i$ have rank at most $r$ (can assume $r=2$). Hence $(A_1, \ldots, A_m)$ live in a projective variety. I interested in the map
$$ (A_1, \ldots, A_m) \mapsto A_1 \odot \ldots \odot A_m$$
where $\odot$ is the Hadamard product.
In general the image of a "morphism" in real algebraic case is semi-algebraic. However, in your case the map is (probably)finite, so proper, hence it has a closed ( in usual topology) image. So it is (probably) defined as a finite union of sets of the form $\{F_1\ge 0, \ldots F_k\ge 0\}$. Explicitely it might get very involved.