Precise definition/construction of $\text{Sym}(\text{Sym}^2)$?
This is clearly explained in Section 6.3 of the reference
[W] Jerzy Weyman, Cohomology of Vector Bundles and Syzygies, Cambridge University Press, Cambridge, 2003.
Ler $X^s$ be the space of $n\times n$ symmetric matrices over a field $\mathbb{K}$. Then the affine coordinate ring $A^{s}=\mathbb{K}[X^s]$ of $X^s$ is isomorphic to a polynomial ring, more precisely by the symmetry condition we have \begin{equation*} A^s \simeq \mathbb{K}[\phi_{ij}]_{\, 1\leq i \leq j \leq n}, \end{equation*} where $\phi_{ij}$ denotes the $ij$-coordinate function on $X^s$.
On the other hand, we may identify $X^s$ with the space $\mathrm{Sym}^2 E^*$, where $E$ is the vector space of dimension $n$ over $\mathbb{K}$, hence we obtain \begin{equation*} A^s \simeq \mathrm{Sym}(\mathrm{Sym}^2E^*). \end{equation*}
This is kind of a silly question to be honest. The name "Sym(Sym^2)" refers to a particular twisted commutative algebra. It doesn't make sense to define what it means without using the word twisted commutative algebra.
In any case, the paper https://arxiv.org/abs/1501.06925v2 of Nagpal-Sam-Snowden defines the twisted commutative algebra $\mathrm{Sym}(\mathrm{Sym}^2(\mathbf C^\infty))$ and proves that it is noetherian. Tom Church's question in Problem 1.5 on the linked page asks if this is true also with $\mathbf Z$-coefficients. I suggest you try to read that paper (and/or other papers like https://arxiv.org/abs/1209.5122), try to understand what a twisted commutative algebra is, and then ask a more focused question.
Francesco's answer is on point, by the way: one possible definition of a twisted commutative algebra is that it is a polynomial functor from vector spaces to commutative rings. The assignment $E^\ast \mapsto A^s$ in Francesco's answer is the polynomial functor that Nagpal-Sam-Snowden prove is noetherian.
As Dan Petersen points out, it's impossible to answer this question without talking about twisted commutative algebras. Moreover, to talk about TCAs over $\mathbb{Z}$, you need a more general definition of TCAs (which becomes equivalent to the earlier Sam-Snowden definitions when you're over $\mathbb{C}$). I'm not sure this is in the literature, so I'll try to explain.
Definition: A TCA is a commutative algebra inside the tensor category of FB-modules.
Let's expand this definition. FB denotes the category of finite sets and bijections, and an FB-module is a functor $W: \mathrm{FB}\to \mathbb{Z}\mathrm{-Mod}$. In other words, $W$ is essentially a collection of $S_n$-representations $W_n$ (one for each $n\in \mathbb{N}$), with no relation between them.
The disjoint union of finite sets $\sqcup:\mathrm{FB}\times \mathrm{FB}\to \mathrm{FB}$ induces a tensor product $\otimes:\mathrm{FB}\mathrm{-Mod}\times \mathrm{FB}\mathrm{-Mod}\to \mathrm{FB}\mathrm{-Mod}$ (often called the "convolution tensor product". It has the universal property that a map of FB-modules $A\otimes B\to C$ is equivalent to a collection of maps $A_S\otimes_{\mathbb{Z}}B_T\to C_{S\sqcup T}$ for all finite sets $S$ and $T$, natural in $S$ and $T$. (Equivalently, you can say it's a collection of $S_k\times S_\ell$-equivariant maps $A_k\otimes_{\mathbb{Z}}B_\ell\to C_{k+\ell}$, but it's easier to keep track of things like the symmetry $A\otimes B\cong B\otimes A$ with the former definition.)
A TCA is an FB-module $A$ together with a multiplication $A\otimes A\to A$ (which has to be commutative, but this is a property not structure so we don't need to stress it).
Now we can define $\mathrm{Sym}(\mathrm{Sym}^2)$. Recall that a perfect matching on a set $T$ is an equivalence relation on $T$ where each equivalence class has exactly 2 elements.
Definition: $\mathrm{Sym}(\mathrm{Sym}^2)$ is the FB-module $A$ which assigns to a finite set $T$ the free abelian group $A_T=\mathbb{Z}[\{\text{perfect matchings on }T\}]$.
To make this a TCA, we need to define the multiplication map $A\otimes A\to A$. According to the universal property above, this means we have to specify natural maps $A_S\otimes_{\mathbb{Z}} A_T\to A_{S\sqcup T}$. But this is very easy: if you have a perfect matching of $S$, and a perfect matching of $T$, their union is a perfect matching of $S\sqcup T$.
From this perspective the name may not be so clear. Here are two explanations. First (and perhaps less satisfying for you?): Define an FB-module $X$ by $X_T=\begin{cases}\mathbb{Z}&\text{if }\lvert T\rvert=1\\0&\text{else}\end{cases}$. Because we have a tensor product, we can make sense of $\textrm{Sym}^k$ in FB-Mod, and for example $\textrm{Sym}^2 X$ turns out to be $(\textrm{Sym}^2 X)_T=\begin{cases}\mathbb{Z}&\text{if }\lvert T\rvert=2\\0&\text{else}\end{cases}$. Moreover, our algebra $A$ is precisely $\textrm{Sym}^*(\textrm{Sym}^2 X)=\bigoplus_{k\in \mathbb{N}}\textrm{Sym}^k(\textrm{Sym}^2 X)$.
Alternately, if you were to work over $\mathbb{C}$ throughout, then there is an alternate description of TCAs over $\mathbb{C}$ in terms of polynomial representations of $\textrm{GL}_\infty(\mathbb{C})$. (See e.g. Section 2.1 of arXiv:1501.06925.) In this perspective, you are looking at nothing other than the commutative ring $\textrm{Sym}^*(\textrm{Sym}^2 \mathbb{C}^\infty)$, with its natural multiplication and natural $\textrm{GL}_\infty(\mathbb{C})$-action.