If we have the symmetric square Sym$^{2}V$ and $V=\mathbb{C}^{2}$, why is it that $\{x^{2}, xy, y^{2}\}$ form a basis for it?
So symmetric square matrices are when the main diagonal acts as a symmetry line i.e for a matrix $A$ the entries would be like $a_{i,j}= a_{j,i}$. I don't think I have much understanding what exactly Sym$^{2}\mathbb{C}^{2}$ is, so any push in the right direction would be helpful thanks
Let me try to shed some light on the different mathematical objects here.
The Symmetric algebra $\operatorname{Sym}(V)$ over a vector space $V$ is a quotient of the Tensor algebra $T(V)$ of that space. If for instance, $V=\{v_1,v_2\}$, then $T(V)$ contains elements like $$v_2,\quad v_1\otimes v_2,\quad v_1\otimes v_2\otimes v_1+v_2\otimes v_2,\text{ etc... }$$
Now, this might look daunting but we can actually interpret these as polynomials in non-commuting variables. Then, if we send $v_1$ to $x$ and $v_2$ to $y$, the terms above become $$y,\quad xy,\quad xyx+y^2 $$
Notice that we write $xyx$ exactly to indicate the non-commutativity. Now, if we understand the tensor algebra via non commuting variables, the symmetric algebra is the correct quotient (of $T(V)$) where the variables commute. That is, in the symmetric algebra, the tensors $v_1\otimes v_2$ and $v_2\otimes v_1$ are the same element (that's why it's called symmetric) and are therefore represented by just $xy=yx$ (now our variables commute...).
In your case, $\operatorname{Sym}^2(V)$ is just the degree $2$ part of the symmetric algebra over $V$. Since that corresponds to polynomials $k[x,y]$, the degree $2$ part is just the homogeneous polynomials of degree $2$, in two variables $x$ and $y$. It should be easy now to see that this is generated by $\{x^2,xy,y^2\}$.
Now, it just so happens that there is a relation between the degree $2$ part of $\operatorname{Sym}(V)$ and symmetric matrices. In the wikipedia article you'll see that one characterization of the degree $k$ part of $\operatorname{Sym}^k(V)$ is that it satisfies a certain universal property, with respect to symmetric multilinear maps defined on $V\times V\times \cdots V$ (product taken $k$ times). When $k=2$, a multilinear map :$V\times V\rightarrow \mathbb{C}$ is exactly an $n\times n$ matrix (where $n=\operatorname{dim}(V)$) and when the multilinear map is symmetric, so is the corresponding matrix.
Therefore, you'll see that the dimension of the space of degree-$2$ homogeneous polynomials in $n$-many variables is equal to the dimension of the space of symmetric $n\times n$ matrices. For example, we can see that all degree-$2$ polynomials on $x,y,z$ are generated by monomials $x^2,y^2,z^2,xy,xz,zy$ and we can see that an arbitrary symmetric $3\times 3$ matrix looks like $\begin{pmatrix}a&d&e\\d&b&f\\e&f&c\\ \end{pmatrix}$ (again, completely specified by $6$ unknowns).
Notice, that if you want to do something similar for $k>2$, you'll have to use objects other than matrices (like $k$-dimension boxes..) and redefine what it means to be symmetric.