Unique symmetric covariant $k$-tensor satisfying $(\operatorname{Sym} T)(A,...,A)=T(A,...,A)$ for all $A \in V$

Here's an analysis of the case $k=2$. Suppose $\bar T$ is another symmetric tensor satisfying $\bar T(A,A)= T(A,A)$ for all $A$.

Now $$T(x+y,x+y)=\bar T(x+y,x+y)= \bar T(x,x)+\bar T(x,y)+\bar T(y,x)+\bar T(y,y).$$ This equals $T(x,x)+2\bar T(x,y)+T(y,y)$. Thus $$\bar T(x,y)=\frac{1}{2}(T(x+y,x+y)-T(x,x)-T(y,y)).$$ So $\bar T$ is uniquely determined by $T$. Since $Sym(T)$ is symmetric and also satisfies the formula $Sym(T)(A,A)=T(A,A)$, $\bar T$ and $Sym(T)$ must be equal.

A similar, but more complicated, argument works in the case of $k>2$.

A symmetric tensor $T:V^n\to k$ descends to a linear functional $\tilde{T}:\mathrm{Sym}^nV\to k$ on the symmetric algebra. Denote the diagonal $V^{[n]}=\{v^n:v\in V\}$. As illustrated in the other answer, when $n=2$,

$$xy=\frac{(x+y)^2-(x^2+ y^2)}{2},$$

demonstrating that every element in $\mathrm{Sym}^2(V)$ can be written as a combination of elements from our set of squares $V^{[2]}$. Similarly, for arbitrary $n$, it suffices for our purposes to show that $\langle V^{[n]}\rangle$ is always $\mathrm{Sym}^nV$, because then every element is a linear combination of the $n$th powers, and so $\tilde{T}$ is uniquely determined by its values on $V^{[n]}$.

Equivalently, we seek to prove that in the polynomial ring $k[x_1,\cdots,x_n]$, the elementary symmetric polynomial $e_n$ can be written as a sum of $n$th powers of degree $1$ homogeneous (but not necessarily symmetric!) polynomials. If we can do this, we can write $x_1\cdots x_n$ as a linear combination of terms from $V^{[n]}$ by mimicking the formula for the polynomial ring.

This is an incomplete answer. I'm not sure if the above thoughts make this question (in the general case) easier, or harder, but it seems like a natural route to follow. I will ask a follow-up question...