Explicit Kodaira-Spencer map of hyperelliptic curves Let $g\geq 2$, and $$\mathcal T=\{(t_1,\cdots,t_{2g+2})~|~t_i\neq t_j,\forall i\neq j\}.$$
For any $t=(t_1,\cdots,t_{2g+2})\in \mathcal T$, let
$$Y_t=\left\{y^2=\prod_{i=1}^{2g+2}(x-t_i)\right\}.$$
Thus we get a family of hyperelliptic curves $\mathcal Y \to \mathcal T$. Let $t^0=(t_1^0,\cdots,t_{2g+2}^0) \in \mathcal T$ be a fixed point (we assume $t_i^0\neq 0,\forall i$). Then we have the Kodaira-Spencer map
$$\Theta_{\mathcal T,t^0} \longrightarrow H^1(Y_{t^0}, \Theta_{Y_{t^0}}).$$
Take dual, we get a map (also called Kodiara-Spencer map):
$$\kappa:H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}}) \longrightarrow \Omega^1_{\mathcal T,t^0}.$$
Now one can take $$\frac{x^i}{y^2}(dx)^2, 0\leq i \leq 2(g-1);~ \frac{x^j}{y}(dx)^2,
0\leq j \leq g-3,$$
as a basis of $H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}})$.
And take $$dt_1, \cdots, dt_{2g+2},$$
as a basis of $\Omega^1_{\mathcal T,t^0}$.
Then can we write down the Kodaira-Spencer map explicitly under the above two basis?
 A: Caution:  I was thinking over the calculation that led to my proposed answer below, and I realized that I had neglected a term that I haven't actually justified as being zero, so now I'm less sure that these formulae are correct.  There may be another term in the formula for $\kappa(\phi_i)$ and I don't see how to rule it out.  As a result, I am withdrawing my answer until I have had time to determine whether the extra term really does vanish.  My apologies to those who trusted my claim and up-voted it!
Unfortunately, I am not allowed to delete an 'accepted answer', so I'm putting this caution at the top so that readers will know not to regard it as correct (yet).

I haven't checked all of the details, but I think that the following is correct and gives you a basis for an answer:  We can get rid of the action of the Möbius group of linear fractional transformations, by, instead, looking at the manifold
$$
\mathcal{S} = \{(s_0,\ldots,s_{2g-2})\  |\   s_i\not=0,1\ \text{and}\ s_i\not=s_j\ \text{when}\ i\not=j \ \}\subset \mathbb{C}^{2g-1}
$$
and letting $s\in\mathcal{S}$ correspond to the genus $g$ hyperelliptic curve $C_s$ defined by
$$
y^2 = x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2}).
$$
The hyperelliptic involution is $\iota(x,y) = (x,-y)$.
A basis for the $\iota$-even holomorphic quadratic differentials on $C_s$ (a space of dimension $2g-1$ when $g>1$) is given by 
$$
\frac{x^i\ dx^2}{x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2})}\ \text{when}\  0\le i\le 2g-2,
$$
and a basis for the $\iota$-odd holomorphic quadratic differentials on $C_s$ (a space of dimension $g-2$ when $g>1$) is given by 
$$
\psi_j = \frac{x^i\ dx^2}{y}\qquad\text{when}\  0\le j\le g-3.
$$
It turns out that a more convenient basis for the $\iota$-even holomorphic quadratic differentials is 
$$
\phi_i = \frac{dx^2}{x(x-1)(x-s_i)}\qquad \text{when}\  0\le i\le 2g-2.
$$
Then, it seems, in this case, that the (dual) Kodaira-Spencer map is given by
$$
\kappa(\psi_j) = 0\qquad \text{when}\  0\le j\le g-3
$$
while there is a (nonzero) constant $c$ (which might depend on $g$, but I don't think so) such that
$$
\kappa(\phi_i) = \frac{c\ ds_i}{s_i(s_i-1)}\qquad \text{when}\  0\le i\le 2g-2.
$$
(This formula even works when $g=1$, by the way.)
I think you can get your desired formula by making this equivariant with respect to the action of the Möbius group $\mathrm{PSL}(2,\mathbb{C})$, but I suspect that the general formula in this case won't be particularly nice.
