deformation theory - Kodaira-Spencer map in a concrete instance

Here is an attempt. Based on your comment to Kevin Lin's post, I think that you know the first part of what I have written, but I included this for the sake of completeness.

• Some Generalities on $phi$: Any deformation of an affine hyperelliptic curve such as

$$
y^2 = prod (x - lambda_i(epsilon))
$$

is trivial and hence corresponds to the zero cohomology class. Indeed, any deformation of a smooth, affine scheme (separated and of finite type over a field?) is trivial. Given a deformation $X_{epsilon} to Delta$ as you describe, the Kodaria-Spencer map is computed by fixing an open affine cover $U_i$ of $X_0$ and isomorphisms $phi_i colon X_{epsilon}|_{U_i} to U_{i} otimes k[epsilon]$ of the restriction of $X_{epsilon}$ to $U_i$ with the trivial deformation of $U_{i}$. The automorphism $phi_{i} circ phi_{j}^{-1}$ of determines an explicit Cech cocyle that represents a class in $H^{1}(X_0, TX_0)$, and one checks that this class is independent of the choices made. The main point: the Kodaira-Spencer class comes from deforming the gluing data NOT from deforming the equations.

• Computation of $phi$: As you wrote, it is not clear from that description how everything works in a concrete cases. Here is how it works out in the case of a general genus $2$ hyperelliptic curve. Working over the field $k$, this curve
  can be described as the curve obtained by gluing the two affine schemes

$$
U_1 := operatorname{Spec}(k[x_1, y_1]/(y_1^2 = prod_{i=1}^{6} (x_1-r_i)),
$$
$$
U_2 := operatorname{Spec}(k[x_2, y_2]/(y_2^2 = prod_{i=1}^{6} (1-r_i x_2)),
$$
over the usual opens via the isomorphism $g$ defined by the rules
$$
x_1 mapsto x_2^{-1},
$$
$$
y_1 mapsto y_2 x_2^{-3}.
$$
Here $r_1, dots, r_6$ are general scalars.

Associated to the affine open cover ${U_1, U_2}$
is the usual Cech complex, and we can use this complex
to compute $H^{1}(X, TX)$. Some elements of this cohomology
group are given by the Cech cocycles
$$
y_1/x_1 frac{partial}{partial x_1}, y_1/x_1^{2} frac{partial}{partial x_1}, y_1/x_1^{3} frac{partial}{partial x_1} in H^{0}(U_{12}, TX).
$$
Here $U_{12}$ denotes the intersection of $U_1$ and $U_2$. Note: one needs to check that these vector fields are regular on $U_{12}$.
The vector field $frac{partial}{partial x_1}$ has simple poles at ramification points of the degree $2$ to $mathbb{P}^1$, and the $y_1$ terms
are needed to cancel these poles. I think these elements form a basis, but
you just asked for an example so I guess we don't care about this.

Let's compute the 1st order deformation of $X$ associated to $D:= y_1/x_1 frac{partial}{partial x_1}$. To construct the deformation, we take
the trivial deformations of $U_1$ and $U_2$ and deform the gluing automorphism. The trivial deformations are
$$
operatorname{Spec}(k[epsilon, x_1, y_1]/(y_1^2 = prod_{i=1}^{6} (x_1-r_i)),
$$
$$
operatorname{Spec}(k[epsilon, x_2, y_2]/(y_2^2 = prod_{i=1}^{6} (1-r_i x_2)).
$$

The general rule is that the deformed gluing map $tilde{g}$ is given by $tilde{g}(a) = g(a) + epsilon cdot g(D(a))$. For our particular choice of
$D$, I think this yields:
$$
x_1 mapsto x_2^{-1} + y_2 x_2^{-2} epsilon,
$$
$$
y_1 mapsto y_2 x_2^{-3} + y_2 x_2^{-2} frac{-x_2^{-1} q'(x_2) + 6 x_2^{-2} q(x_2)}{2 y_2} epsilon.
$$
Here $q(x_2) = prod_{i=1}^{6} (1-r_i x_2)$.

The expression for the image of $y_1$ is quite complicated, but it hopefully is just
$g(y_1/x_1 frac{partial y_1}{partial x_1})$.

One can work our a similar description for the deformations coming from the other cohomology classes that I wrote down. Assuming these form a basis, this completely describes the map $phi$.

It is easy to reverse this construct as well. Every deformation arises by deforming the map $g$ to a map $tilde{g}$ as we have done. The associated cohomology class can be described by writing $tilde{g} = g + epsilon cdot D$ for some function $D$. One can show that $D$ defines a regular vector field on $U_{12}$ and hence represents an element of $H^{1}(X, TX)$.