Lie algebra action from Lie group action: coordinates

I will try my best not to give away everything. I will use your notation above: $\pi$ is the representation of $SL_2(\Bbb{C})$ that you defined above and we have the induced representation $d\pi$ on the Lie algebra $\mathfrak{sl}_2(\Bbb{C})$. Recall that

$$d\pi(X) = \frac{d}{dt}\pi(e^{tX})\bigg|_{t=0}.$$

From which we get that if $f \in V_d$, $(d\pi(X)f)(z,w) = \frac{d}{dt}f(e^{-tX}z,e^{-tX}w)\bigg|_{t=0}.$ You now have a formula for computing the action of any $X \in \mathfrak{sl}_2(\Bbb{C})$ on any $f \in V_d$. Now because the usual matrices $H,E,F$ form a basis for $\mathfrak{sl}_2(\Bbb{C})$ you only need to compute the matrices for $d\pi(E),d\pi(F),d\pi(H)$.

Now you mentioned in you naive approach above that you want to compute $\pi$ of any matrix in $\mathfrak{sl}_2(\Bbb{C})$ on the canonical basis for $V_d$. Instead, what I suggest you try is instead of working on basis elements for $V_d$ just work with a general homogeneous polynomial $f \in V_d$.

As a start, from the formula that I gave you above you can use the chain rule to simplify things. Try working out the action of $H$ in the Cartan subalgebra on a general $f \in V_d$. You should get that

$$(d\pi(H)f)(z,w) = - \frac{\partial f}{\partial z} z + \frac{\partial f}{\partial w}w$$

from which it follows that $d\pi(H)$ is the linear operator on $V_d$ given by.....

Even though this question has been answered, let me add something more to your general question. The general result you inquire about is true even under less restrictive circumstances. Suppose $N$ is a subset of $M$ which has a smooth structure (not necessarily one induced by the smooth structure on $M$) and such that the inclusion map $i_N:N\hookrightarrow M$ is smooth. Then for $f:M\rightarrow Y$ smooth, $d(f|_N) = d(f\circ i_N) = df\circ di_N$, implying that $d(f|_N):TN\rightarrow TY$ is determined by $df:TM\rightarrow TY$. In your case, since $0$ is a regular point of $F$, the smooth structure on $N=F^{-1}(0)$ induced by that on $M$ makes $N$ an embedded submanifold of $M$, with tangent space $TN \simeq di_N(TN) = \ker dF$, as you say.