Your first question can also be answered using the symmetric algebra $\bigvee V^*$, where $V^*$ is the dual space of $V$. The result follows from:
Theorem: If $z^*\in\bigvee^p V^*$ and $i(v)^pz^*=0$ for all $v\in V$, then $z^*=0$.
Here $i(v):\bigvee V^*\to\bigvee V^*$ is the insertion operator dual to the multiplication operator $v\vee(-):\bigvee V\to\bigvee V$.
Proof: By induction on $p$. If $p=0$, the result is trivial. If $p\ge 1$ and the result holds for $p-1$, fix $w\in V$ and consider $i(w)z^*\in\bigvee^{p-1}V^*$. For $v\in V$, define
$$f(\lambda)=i(\lambda v+w)^pz^*=[\lambda i(v)+i(w)]^pz^*=\sum_{r+s=p}\frac{p!}{r!\,s!}\,\lambda^r\,i(v)^ri(w)^sz^*$$
By assumption $f(\lambda)=0$ for all $\lambda\in\mathbb{R}$, so it follows that $i(v)^ri(w)^sz^*=0$ for all $r,s$ with $r+s=p$, in particular
$$i(v)^{p-1}i(w)z^*=0$$
Since $v$ was arbitrary, it follows from the induction hypothesis that $i(w)z^*=0$, and this holds for any $w$ since $w$ was arbitrary. Now if $z\in\bigvee^p V$ we can write $z=\sum_i w_i\vee z_i$ for $w_i\in V$ and $z_i\in\bigvee^{p-1}V$. It follows that
$$\langle z^*,z\rangle=\sum_i\langle z^*,w_i\vee z_i\rangle=\sum_i\langle i(w_i)z^*,z_i\rangle=0$$
under the duality between $\bigvee^p V^*$ and $\bigvee^p V$, and since $z$ is arbitrary it follows that $z^*=0$.
In your case, you can apply this result to $S-T$ to conclude that $S-T=0$.
