Why is the group action on the vector space of polynomials naturally a left action?

The action defined here is on functions, i.e. $P$, not vectors $z$. Write $R=g_2\cdot P$. Then $(g_1\cdot R)(z)=R(zg_1)$. Combine this with the fact that $R(x)=P(xg_2)$ for any vector $x$. Here $x=zg_1$, so we get that $$ (g_1\cdot(g_2\cdot P))(z)=(g_1\cdot R)(z)=R(zg_1)=(g_2\cdot P)(zg_1)=P((zg_1)g_2). $$

Another way of seeing this is to realize that $g\cdot P$ is the function gotten by composing $\rho_g:z\mapsto zg$ with $P$ from the right (i.e. ahead of $P$). So $$ g\cdot P = P \circ \rho_g. $$ Therefore $$ g_1\cdot(g_2\cdot P)=g_1\cdot(P\circ \rho_{g_2})=(P\circ \rho_{g_2})\circ\rho_{g_1}. $$ By associativity of composition this latter composition is then $$ P\circ(\rho_{g_2}\circ\rho_{g_1})=P\circ\rho_{g_1g_2}=(g_1g_2)\cdot P. $$ Here $\rho_{g_2}\circ\rho_{g_1}=\rho_{g_1g_2}$ because $$(\rho_{g_2}\circ\rho_{g_1})(z)=\rho_{g_2}(\rho_{g_1}(z))=\rho_{g_2}(zg_1)=(zg_1)g_2=z(g_1g_2)=\rho_{g_1g_2}(z). $$

First, you have a left action on vectors, $g \cdot z = zg^{-1}$. (Here $g$ is an invertible matrix, $z$ is a row vector and $g \cdot z = zg^{-1}$ is the result of multiliplying $z$ times $g^{-1}$. You need to multiply on the right by $g^{-1}$ to make it a left action.) Then $(g \cdot P)(z) = P(g^{-1} \cdot z) = P(zg)$.

Let's check that this is a left action, i.e. $g_1 \cdot (g_2 \cdot P) = (g_1 g_2) \cdot P$: $$ (g_1 \cdot (g_2 \cdot P) )(z) = (g_2 \cdot P)(z g_1) = P(z g_1 g_2) = ((g_1 g_2) \cdot P)(z). $$

The reason that the action is given by $(g \cdot P)(z) = P(g^{-1} \cdot z)$ is because the polynomials on $V$ are the elements of the symmetric algebra of $V^*=\text{Hom}(V,\mathbb{C})$. Given a left $G$-action on $V$ (i.e. a representation), there is a natural way to construct a left $G$-action on $V^*$ -- the contragredient representation -- by letting $(g \cdot l)(v) := l(g^{-1} \cdot v)$. The argument for why this defines an action on $V^*$ follows from the calculation above. (Indeed, just let $P = l$ be linear.) Similarly, there is also a natural way to get a $G$-representaion of $Sym^{\bullet}(V^*)$ from a $G$-representation of $V$, which yields the action on the polynomial ring that you are interested in.