So I'm studying the lecture notes of Wendl on Holomorphic curves in symplectic geometry and I try to understand the linearization of $\bar\partial_J$. My question is, suppose we have a suitable vector bundle $E \to B$ and a section $s:B\to E$, why is the linear map $\nabla s: T_pB \to E_p$ independent of the choice of $\nabla$ at any point $p$ where $s(p)=0$ ?
I was told, this is because the difference of two connections is $C^\infty(B)$-linear, but I don't quite understand this argument.
By the way all of this can be found on page 44 of Wendls notes. I'm not sure if I'm allowed to post a link, but you can find them easily.
Given two connections $\tilde{\nabla},\nabla$, you can define $\omega \colon \mathcal{X}(B) \times \Gamma(E) \rightarrow \Gamma(E)$ by $\omega(X,s) = \tilde{\nabla}_X(s) - \nabla_X(s)$. You can verify using the product rule for connections that $\omega$ is $C^{\infty}(B)$-linear with respect to both variables. By a general useful fact in differential geometry, this implies that $\omega$ can actually be interpreted as a pointwise linear map between vector bundles. Namely, there is a smooth section $\overline{\omega} \in \Omega^1(B,\operatorname{End}(E))$ such that $$ \omega(X,s)(p) = \left( \overline{\omega}|_{p}(X_p) \right)(s_p). $$ This implies that if $s(p) = 0_p$ then $$ \tilde{\nabla}_X(s)|_{p} - \nabla_X(s)|_{p} = \omega(X,s)(p) = \left( \overline{\omega}|_{p}(X_p) \right)(0_p) = 0.$$
Alternatively, you can verify this fact using a calculation in coordinates. Choose some neighborhood $U$ of $p$ in $B$ over which $E$ is trivial with sections $s_1,\dots,s_k$ and define the connection forms $\omega,\tilde{\omega}$ by $$ \nabla_{X}(s_i) = \omega_i^j(X) s_j, \tilde{\nabla}_{X}(s_i) = \tilde{\omega}_i^j(X) s_j.$$
Given a section $s$ with $s(p) = 0$ write $s = h^i s_i$ over $U$ so that $h^1(p) = \dots = h^k(p) = 0$. Then
$$ \tilde{\nabla}_X(s) = \tilde{\nabla}_X(h^i s_i) = d(h^i)(X) s_i + h^i \tilde{\omega}_i^j(X) s_j, \\ \nabla_X(s) = d(h^i)(X) s_i + h^i \omega_i^j(X) s_j, \\ \tilde{\nabla}_X(s) - \nabla_X(s) = h^i \left( \tilde{\omega}_i^j(X) - \omega_i^j(X) \right) s_j $$
and since $h^i(p) = 0$ we see directly from the formula that $$ \tilde{\nabla}_X(s)|_{p} = \nabla_X(s)|_{p}. $$
This calculation also shows that the difference between two connections is actually tensorial and given in local coordinates by the difference between the two connections forms.