I would like to understand deeply the relationship between the symmetric tensor product of order 2 and the sum of squares.
For me, it is clear that the symmetric tensor product or order $d$ is isomorphic to the set of the homogeneous polynomials of degree $d$. I want to know precisely if it is true the following statement:
If $V$ is a vector space of monomials, then $Sym^2(V)$ is the space of the polynomials that can be written as a sum of squares?
No, it's not true. First of all, if $V$ is a vector space spanned by monomials of some degree $d$, then $V$ is a vector space of (homogeneous) polynomials (of degree $d$). Then $\operatorname{Sym}^2(V)$ is not identified with the polynomials of $2d$, for example $(x^2)(yz)$ and $(xy)(xz)$ are different elements of $\operatorname{Sym}^2(\operatorname{span}\{x^2,xy,\dotsc,z^2\})$, although of course they map to the same polynomials. (In general if $V = \operatorname{Sym}^d(W)$, then $\operatorname{Sym}^e(V) = \operatorname{Sym}^e(\operatorname{Sym}^d(W))$ maps naturally to $\operatorname{Sym}^{de}(W)$ but it's not an isomorphism.)
Finally $\operatorname{Sym}^2(V)$ is the space of quadratic forms in a basis of $V$, but I'm not sure if I would agree with the description as polynomials that can be written as sums of squares. They aren't polynomials in whatever variables form the monomials spanning $V$, and more importantly there's no positivity condition. For example $\operatorname{Sym}^2(V)$ includes elements like $(xy)^2 - (xz)^2$, a linear combination of squares; if you're okay with calling that a "sum", then it's true that every quadratic form is a sum, or linear combination, of squares (but again these are not exactly quadratic forms). If you want sums with positive coefficients, then you are looking for a cone in the vector space of quadratic forms.
To try to be clear: if $V = \operatorname{span}\{x_1,\dotsc,x_n\}$ is spanned by variables, then $\operatorname{Sym}^2(V)$ can be identified with quadratic forms in the $x_i$ (as long as we're not in characteristic $2$), and every element of $\operatorname{Sym}^2(V)$ is a linear combination of squares, e.g., $x_i x_j = (1/4)((x_i+x_j)^2-(x_i-x_j)^2)$. But if $V = \operatorname{Sym}^d(\operatorname{span}\{x_1,\dotsc,x_n\})$ is already the space of degree $d$ polynomials in the $x_i$, then $\operatorname{Sym}^2(V)$ is not the space of degree $2d$ polynomials in the $x_i$, it's just a space with a natural surjection onto that.
