As you write, the geometric structure factor is
$$F_{hkl} = \sum_{j=1}^N f_j e^{i \delta k \cdot \vec r_j}$$
You also correctly state that $\vec{r}_j$ denotes the location of the $j$-th atom in the cell.
Now, you incorrectly say that $\vec{r}_2 = \vec{a}_1$, $\vec{r}_3 = \vec{a}_2$.
But $\vec{a}_1$ doesn't tell you where in the cell the 1st atom is, $\vec{a}_1$ tells you in what direction the next cell starts. The $\vec{a}_i$ are called the lattice vectors and are just there to define the cubic structure. They can be chosen the same for the simple cubic, the bcc and the fcc lattice.
What you need are the basis vectors. In a simple cubic lattice, you only have one basis vectors, $\vec{0}$. In the fcc lattice, we have four atoms per unit cell, and therefore we have four basis vectors, and those are the vectors that you have written down above: $\vec{0}, a/2 (\vec{x}+ \vec{y})$ and so on, where $\vec{x}$ is the unit vector in $x$-direction. In a cubic lattice, we have
$\vec{a}_1 = a\vec{x}, \vec{a}_2 = a\vec{y}, \vec{a}_3 = a \vec{z}$, so your basis vectors read $\vec{r}_1 = \vec{0}, \vec{r}_2 = 1/2(\vec{a}_1 + \vec{a_2})$ etc.
Now if you insert those into the exponential and use the fact that $\vec{a}_i \cdot \vec{b}_j = 2\pi\delta_{ij}$ you should get the correct result.
Best Answer
You don't need to. In fact, you can work directly in 2D and solve things explicitly, since the condition for the reciprocal basis that $b_i\cdot a_j = 2\pi\delta_{ij}$ reads in matrix notation $$ \begin{pmatrix} b_{1x} & b_{1y} \\ b_{2x} & b_{2y} \end{pmatrix} \begin{pmatrix} a_{1x} & a_{2x} \\ a_{1y} & a_{2y} \end{pmatrix} = 2\pi \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , $$ so all you need to do is multiply with the explicit matrix inverse on the right to get $$ \begin{pmatrix} b_{1x} & b_{1y} \\ b_{2x} & b_{2y} \end{pmatrix} = \frac{2\pi}{a_{1x}a_{2y}-a_{1y}a_{2x}} \begin{pmatrix} a_{2y} & -a_{2x} \\ -a_{1y} & a_{1x} \end{pmatrix} , $$ or, transposing this for clarity so that you can read off the matrix equation column-by-column to get the $b_i$, $$ \begin{pmatrix} b_{1x} & b_{2x} \\ b_{1y} & b_{2y} \end{pmatrix} = \frac{2\pi}{a_{1x}a_{2y}-a_{1y}a_{2x}} \begin{pmatrix} a_{2y} & -a_{1y} \\ -a_{2x} & a_{1x} \end{pmatrix} . $$ You can then check by hand that this matches the projection on the $x,y$ plane of the results you get from the 3D formulas by taking the third vector along $z$.
If, on the other hand, you do want to stick to the three-dimensional formalism (which then lets you use the same formulas for both cases) then you need to supplement your basis of the plane with a third vector to make the span three-dimensional. This third vector needs to have a nonzero $z$ component (or it would be linearly dependent on $a_1$ and $a_2$), but you need additional restrictions to specify it uniquely.
These restrictions come from the fact that
As to the precise magnitude of $a_3$, it is irrelevant ─ it is easy to see that changing $a_3\mapsto \lambda a_3$ by any $\lambda \neq 0$ does not affect in any way the 3D $b_1$ and $b_2$ expressions you quote.