Ah, I see what you mean now. You are absolutely right! The statement I made about $\theta_{x}$ vanishing is incorrect. We can have a Higgs field $\theta$ with a multiple zero of the determinant at $x$ for which nevertheless the $\theta_{x}$ is a $2\times 2$ Jordan block. This happens when $\theta$ parametrizes a family of regular semisimple elements near $x$ specializing to a regular nilpotent element at $x$. I will correct the answer above to reflect this.

I did not understand your last comment. If the the spectral cover is of the form $y^{2} = a^{2}$, then $\det(\theta) = - a^{2}$ so all zeroes are double.

Since your $\theta$ is traceless the corresponding specral cover has equation $y^{2} - \det(theta) = 0$. So $\det(\theta)$ vanishes exactly at the branch points of the spectral cover. The determinant vanishes to first order at $x$ if the spectral cover is smooth and branched over $x$. This happens precisely when $\theta_{x}$ has a single eigenvector, i.e. when it has one $2\times 2$ Jordan block. The determinant vanishes to second order if and only if the spectral cover has a singularity at $x$, i.e $\theta_{x}$ has two eigenvectors , i.e. $\theta_{x} = 0$.

$Tot(\mathcal{O}_{\mathbb{P}^{n}}(k))$ is $\mathbb{P}^{n+1}(1,1,\ldots,1,k) - \{x\}$, where $x = (0:0:\ldots:0:1)$.

Nice example with the Fano variety of lines! I still don't understand why you say that the answer of the second question is 'yes'? Your example shows that the answer to the second question is 'no'. and I think my elementary example with a non-linear family of curves on a K3 (in the comment above) also shows that the answer is 'no'. The first one has a chance of being true though.

Still, I am not sure why it is expected that the first order deformation determines the intersections. What if we take a linear pencil of high genus curves in a K3, and then take a curve $\Lambda$ in the parameter space of the linear system which is tangent to the line parameterizing the pencil at some point $p$. Then the divisors corresponding to points $t \in \Lambda$ near $p$ will intersect the divisor corresponding to $p$ at a locus that varies with $t$ and has nothing to do with the pencil.

Do you have some hidden compactness assumption here? What if you take $X$ to be the cotangent bundle of the complex line (with coordinate $x$), $Y$ to be the zero section and and $Y_{\gamma(t)}$ to be the graph of the closed one form $t(x-t)dx$. In this case the section $s$ is $xdx$ so it vanishes at $x=0$. But $Y_{\gamma(t)}$ intersects $Y$ at the point $x =t$.