Intuitively, you can see ot this way: for any closed subset $Y\subset X$, there is a maximal sheaf of quasi-coherent sheaves defining $Y$: this is the reduced structure on $Y$. This sheaf $\mathcal{I}$ is precisely the sheaf of functions vanishing on $Y$.
If $U = X\setminus Y$, this condition is empty on $U$, and we get $\mathcal{I}_{|U} = {O_X}_{|U}$.
More generally, if $\mathcal{I}$ is any quasi-coherent sheaf defining $Y$, we just need to show that $\forall x \in U$, $1\in \mathcal{I}_x$. But saying that $x$ is not in $Y$ is the same as saying that there is a function $f$ of $ \mathcal{I}$ defined on a neighbourhood of $x$, which does not vanish at $x$. This function is invertible in the stalk $O_{X,x}$, so $ \mathcal{I}_x = O_{X,x}$.
Honestly, I have no idea what you are on about connections and such…
Let $(a_{i,j})$ be an arbitrary $7\times 3$ matrix of scalars, and let $f_1,\dots,,f_7$ be the linear combinations of monomials indicated in the rows of following table:
$$\begin{array}{*{12}{c}}
x^2 & y^2 & z^3 & w^2 & xy & xz & xw & yz & yw & zw \\ \hline
1 & & & & & & & a_{1,1} & a_{1,2} & a_{1,3} \\
& 1 & & & & & & a_{2,1} & a_{2,2} & a_{2,3} \\
& & 1 & & & & & a_{3,1} & a_{3,2} & a_{3,3} \\
& & & 1 & & & & a_{4,1} & a_{4,2} & a_{4,3} \\
& & & & 1 & & & a_{5,1} & a_{5,2} & a_{5,3} \\
& & & & & 1 & & a_{6,1} & a_{6,2} & a_{6,3} \\
& & & & & & 1 & a_{7,1} & a_{7,2} & a_{7,3} \\
\end{array}$$
Now let $a$, $b$, $c$ $d$ be four scalars and consider the ideal generated by the $7$ polynomials
$$
f_1(x-a,y-b,z-d,w-d), \dots, f_7(x-a,y-b,z-d,w-d)
$$
and all the polynomails $(x-a)^i(y-b)^j(z-c)^k(w-d)^l$ with $i+j+k+l=3$.
This gives you a $25$-dimensional family of ideals of colength $8$, parametrized by a point in $k^4\times M_{7,2}(k)$.
Viewing the entries of the matrix and the coordinates of the point $(a,b,c,d)$ as varibles now, the ideal generated by those seven polynials in $k[x,y,z,w,a,b,c,d,a_{1,1},\dots,a_{7,3}]$ define subscheme $Z$ in $k^4\times M_{7,2}(k)\times k^4$. The restriction of the map $p:k^4\times M_{7,2}(k)\times k^4\to k^4\times M_{7,2}(k)$ projecting on the first two factors to $Z$ is a map $Z\to k^4\times M_{7,2}(k)$ which is a flat family of subschemes of $k^4$, the fiber of $p$. The universal property of the Hilbert scheme tells you then that to this flat family corresponds a regular map into the Hilbert scheme.
Best Answer
As a module over $D$, $k = D/(t)$ so tensoring with it means killing $t$. This is exactly the assumption the previous sentence.
In more detail, you can apply the tensor with $k$ to the sequence $I’ \to B’ \to B’/I’$ and use right exactness and $B’\otimes k = B$ and $I’\otimes k = I$.