I also know that given the length of 2 sides in a kite and the angle of one of the other angles (which aren't included angles), you can find the area by multiplying these two sides and the sine of the angle. I am unable to find a relationship between this situation and the situation posted in the title question.
Also I can't visualize why the situation in the title question is true. Thanks!
On
The quadrilateral consists of two triangles of the same base $d_1$. Therefore the area is
$$A=\frac12h_1b_1+ \frac12h_2b_1=\frac12(h_1+h_2)d_1$$
Let the two segments of $d_2=x+y$ and recognize that $h_1=x \sin\alpha $ and $h_2=y \sin\alpha $. Then,
$$A = \frac12 (x\sin\alpha+y\sin\alpha)d_1=\frac12(x+y)\sin\alpha d_1=\frac12d_1d_2\sin\alpha$$
In fact, the area of a quadrilateral is one-half the product of its diagonals times the sine of their included angle.
The area of this triangle is $\frac12bh$. $\sin\alpha=\frac ha$, so $h=a\sin\alpha$ and thus the area of the triangle can also be expressed as $\frac12ab\sin\alpha$.
In your quadrilateral, let the two segments of $d_1$ have lengths $w$ and $x$ and the two segments of $d_2$ have lengths $y$ and $z$. We will use the formula above to find the sum of the four triangles in the quadrilateral. Note that two of those two triangles have an included angle measure of $\alpha$ and the other two have the supplement of $\alpha$, but we will ignore that because both of those angles have the same sine. Therefore, the area of the quadrilateral is $$\frac12wy\sin\alpha+\frac12wz\sin\alpha+\frac12xy\sin\alpha+\frac12xy\sin\alpha\\=\frac12(wy+wz+xy+xz)\sin\alpha\\=\frac12(w+x)(y+z)\sin\alpha=\frac12d_1d_2\sin\alpha$$