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 $H^{1}(X, TX)$.
