Assume that $C$ is a curve with a singular point $c_0$. I would like to know about the normal sheaf $\mathcal N_{Z/C}$, where $Z$ is a length two cycle entirely supported at $c_0$ and about $H^1(Z,\mathcal N_{Z/C})$.
This question arises, since I would like to compute the tangent space to $\operatorname{Hilb}^2(C)$ at the points corresponding to such a $Z$. I saw some computations of this but only in the case when everything was smooth and the normal sheaf is a bundle, not just a sheaf. Here, I don't even get any idea of how the normal sheaf should look like or how to compute the $H^1$. Please help/give hints!
Let me give the simplest example. Let $C \subset \mathbb{A}^2$ be given by the equation $xy = 0$, and let $Z \subset C$ be the subscheme of length $2$ defined by equations $$ y - ax = x^2 = 0. $$ I will assume $a \ne 0,\infty$ (so that the line $y - ax = 0$ is not a component of $C$). Then $Z$ is the scheme-theoretic intersection of $C$ with the line $y - ax = 0$ (because there is an equality of ideals $(y-ax,xy) = (y-ax,ax^2) = (y-ax,x^2)$). This means that $Z \subset C$ is a Cartier divisor with equation $y - ax = 0$, hence its structure sheaf has a resolution $$ 0 \to O_C \stackrel{y-ax}\to O_C \to O_Z \to 0. $$ In the other words, $I_Z \cong O_C$. It follows that the conormal sheaf is given by $$ I_Z/I_Z^2 \cong I_Z \otimes_{O_C} O_Z \cong O_C \otimes_{O_X} O_Z \cong O_Z. $$
Now, what you are really interested in is not $H^i(Z,N_{Z/C})$, but rather $Ext^i(I_Z,O_Z)$ (in this particular case these are the same, but in general these spaces may be different). In this particular case $I_Z = O_C$ implies that $$ \mathcal{H}om(I_Z,O_Z) \cong O_Z, $$ and $$ \mathcal{E}xt^p(I_Z,O_Z) = 0 $$ for $p > 0$ since $I_Z$ is locally free (hence locally projective). From the local-to global spectral sequence $$ H^q(C,\mathcal{E}xt^p(I_Z,O_Z)) \Rightarrow Ext^{p+q}(I_Z,O_Z) $$ we conclude $$ Ext^q(I_Z,O_Z) = H^q(C,\mathcal{H}om(I_Z,O_Z)) = H^q(C,O_Z) $$ which is equal to $k[x,y]/(y-ax,x^2) \cong k^2$ for $q = 0$ and vanishes for $q > 0$.