Your first equation, which is a mere definition for the current, applies to regular vector fields $\mathbf J$. It is not straightforward to apply it to Dirac delta-functions. If you consider a finite-size wire, when it is angled wrt $S$ the intersection is larger by a factor $1/\cos θ$ which compensates exactly the $\cos θ$ factor arising from the dot product.
Now for your second question. Let's start by writing the current density correctly for an infinitesimally thin wire placed along the $z$ axis (it is defined by equations $x=0, y=0$, hence the two delta-functions):
$$\mathbf j=I δ(x)δ(y)\hat z.$$ Suppose the surface of integration $S$ is a rectangle in a plane $(x',y)$ that's tilted by an angle $θ$ with respect to the $(x,y)$ plane. Viewed from the side:
The normal $\mathbf n$ (along a $z'$ axis angled by $θ$ wrt $z$) has coordinates $(\sin θ,0,\cos θ)$ in the $(x,y,z)$ system. So
$$\iint_S \mathbf j·\mathbf n\,\mathrm dS=\iint_S Iδ(x)δ(y)\cos θ\,\mathrm dS.$$
Then we have to express the delta-functions in $x'y$ coordinates. $y$ is unchanged. Equation $x=0$ becomes $x'\cos θ+z'\sin θ=0$, so $δ(x)$ is replaced by $δ(x'\cos θ+z'\sin θ)$. On $S$, $z'$ is always zero, so finally we have to calculate
$$\iint_S Iδ(x'\cos θ)δ(y)\cos θ\,\mathrm dS.$$
Now, you should know $δ(k x')=\frac 1{|k|}δ(x')$, hence
$$\iint_S Iδ(x'\cos θ)δ(y)\cos θ\,\mathrm dS=\iint_S Iδ(x')δ(y)\,\mathrm dS=I.$$
(Alternatively, express $\mathrm dS=\mathrm dx'\,\mathrm dy$ and substitute variable $x''=x'\cos θ$.)
Best Answer
Ampere's law (for a steady current) states that $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I$$
If we consider an infinite wire, then symmetry tells us that the B-field at the point $A$ and all other points on a circle of radius $(R+y)$ is constant in magnitude and is in the azimuthal direction. Hence the magnitude of the B-field is given by $$ 2 \pi (R+y)B = \mu_0 I$$ $$ B = \frac{\mu_0 I}{2\pi(R+y)}$$
So now, for a semi-infinite wire, I take away half the wire and hence half the vector field. But, before I just say that the new field is half of the original one, I need to establish that the B-fields from each "half" of the infinite wire are in fact in the same direction so that they add in parallel fashion. The Biot-Savart law tells us that each wire element produces a B-field that is perpendicular to the current and perpendicular to a displacement joining the wire element and the point at which I wish to know the field. So, the B-field is always in a direction azimuthal to the wire, whichever piece(s) of wire we consider. This means that the new B-field for the semi-infinite wire is in the same direction as for the full infinite wire but has half the magnitude.