I've come to the conclusion that Voisin deliberately includes the sign for $\alpha_t$ to make Proposition 9.7 work:
The map $T_{B,0} \to A^{0,1}(T_X)$ given by $u \mapsto d_u(\alpha_t)$ has values in the set of $\overline{\partial}$-closed sections of $\mathcal A^{0,1}(T_X)$ and for $u \in T_{B,0}$, the Dolbeault cohomology class of $d_u(\alpha_t)$ in $H^1(X, T_X)$ is equal to $\rho(u)$.
Here $\rho$ is the Kodaira-Spencer map $T_{B,0} \to H^1(X, T_X)$ obtained by the long exact sequence associated to
$$0 \to T_X \to T_{\mathcal X}|_X \to \phi^* T_B|_X \to 0.$$
This is not a very natural isomorphism, I guess. I'm not sure where you found this. One is a real vector space and the other is a complex vector space, so in what sense are these isomorphic? I will give the argument to deduce that the two have the same dimension as real vector spaces. I will use the Dolbeault isomorphism $H^{p,q}(M) \cong H^q(M,\Omega^p)$.
If the genus of $M$ is $g$, then $\dim_{\Bbb R} H^1(M,\Bbb R) = 2g$ and $g=\dim_{\Bbb C} H^0(M,\Omega^1) = \dim_{\Bbb C} H^{1,0}(M)$. On the other hand, it follows from harmonic theory (taking complex conjugates of harmonic representatives) that $H^1(M,\mathscr O) \cong H^{0,1}(M) \cong \overline{H^{1,0}(M)}$, and so $\dim_{\Bbb C} H^1(M,\mathscr O) = g$, as well.
You can deduce the statement slightly more indirectly from the Hodge decomposition: $H^1(M,\Bbb C) \cong H^{1,0}(M)\oplus M^{0,1}(M)$. Then
$$H^1(M,\Bbb R)\otimes\Bbb C \cong H^1(M,\mathscr O)\oplus \overline{H^1(M,\mathscr O)}.$$
The claim on dimensions follows immediately from this.
Best Answer
Recall that, in general, $H^{p,0}(X)=H^0(X,\Omega^p_X)$. Now $\Omega_\mathbb C^1$, the co-tangent bundle, is the trivial bundle of rank $1$ on $\mathbb C$, the holomorphic sections of which are just the holomorphic functions, of which there are plenty.