Molecular orbital diagram for BF3
I think you are most of the way to the answer, but I will start the process from scratch for the sake of a full explanation for future readers.
First, we determine the symmetry group of $\ce{BF3}$, which by quick inspection we can determine to be $D_{3h}$ (since it has a principle $C_3$ axis, $3\perp C_{2}$ axes, and a horizontal mirror plane of symmetry).

Knowing this, we can now make use of the $D_{3h}$ character table (thank you @orthocresol for compiling these in mathjax format) to determine the irreducible representation of the atomic orbitals (or their symmetry adapted linear combinations SALCS) within the group.
$$\begin{array}{|c|cccccc|c|c|} \hline
D_\mathrm{3h} &amp; E &amp; 2C_3 &amp; 3C_2 &amp; \sigma_\mathrm{h} &amp; 2S_3 &amp; 3\sigma_\mathrm{v} &amp; \text{linear/rotations} &amp; \text{quadratic}\\ \hline
\mathrm{A_1'} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; &amp; x^2+y^2,z^2 \\
\mathrm{A_2'} &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; R_z &amp; \\
\mathrm{E'} &amp; 2 &amp; -1 &amp; 0 &amp; 2 &amp; -1 &amp; 0 &amp; (x,y) &amp; (x^2-y^2,xy) \\
\mathrm{A_1''} &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; &amp; \\
\mathrm{A_2''} &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; 1 &amp; z &amp; \\
\mathrm{E''} &amp; 2 &amp; -1 &amp; 0 &amp; -2 &amp; 1 &amp; 0 &amp; (R_x,R_y) &amp; (xz,yz) \\ \hline
\end{array}$$
Using this, we can quickly determine the irreps of boron's orbitals using the 2nd column from the right.

$\ce{B}$: $2s=a_1'$, $2p_x,2p_y=e'$, and $2p_z=a_2'$.

Next, we will determine the irreps for the $3\ce{F}$ atoms, which will require that I consider combinations of $s,p_x,p_y, \text{and } p_z$ orbitals.
Below, I show how the symmetry operations of the group affect each set of orbitals and show the orbitals that result (image source).
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} &amp; E &amp; 2C_3 &amp; 3C_2 &amp; \sigma_\mathrm{h} &amp; 2S_3 &amp; 3\sigma_\mathrm{v} &amp; &amp; \\ \hline
\Gamma_s &amp; 3 &amp; 0 &amp; 1 &amp; 3 &amp; 0 &amp; 1 \\ \hline
\mathrm{A_1'} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\
\mathrm{E'} &amp; 2 &amp; -1 &amp; 0 &amp; 2 &amp; -1 &amp; 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} &amp; E &amp; 2C_3 &amp; 3C_2 &amp; \sigma_\mathrm{h} &amp; 2S_3 &amp; 3\sigma_\mathrm{v} &amp; &amp; \\ \hline
\Gamma_{p_x} &amp; 3 &amp; 0 &amp; 1 &amp; 3 &amp; 0 &amp; 1 \\ \hline
\mathrm{A_1'} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\
\mathrm{E'} &amp; 2 &amp; -1 &amp; 0 &amp; 2 &amp; -1 &amp; 0  \\ \hline
\end{array}$$
$\hspace{12ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} &amp; E &amp; 2C_3 &amp; 3C_2 &amp; \sigma_\mathrm{h} &amp; 2S_3 &amp; 3\sigma_\mathrm{v} &amp; &amp; \\ \hline
\Gamma_{p_y} &amp; 3 &amp; 0 &amp; -1 &amp; 3 &amp; 0 &amp; -1 \\ \hline
\mathrm{A_2'} &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 \\
\mathrm{E'} &amp; 2 &amp; -1 &amp; 0 &amp; 2 &amp; -1 &amp; 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} &amp; E &amp; 2C_3 &amp; 3C_2 &amp; \sigma_\mathrm{h} &amp; 2S_3 &amp; 3\sigma_\mathrm{v} &amp; &amp; \\ \hline
\Gamma_{p_z} &amp; 3 &amp; 0 &amp; -1 &amp; -3 &amp; 0 &amp; 1 \\ \hline
\mathrm{A_2''} &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; 1 \\
\mathrm{E''} &amp; 2 &amp; -1 &amp; 0 &amp; -2 &amp; 1 &amp; 0  \\ \hline
\end{array}$$
$\hspace{21ex}$
These reductions to irreducible representations can be done via the reduction formula $$n_i=\frac{1}{h}\sum_c g_c \cdot \chi_{i_c} \cdot \chi_{r_c}$$ where $n_i$ is the number of times a particular irrep $i$ occurs in the reducible representation, $g_c$ is the number of symmetry operations $c$, $\chi_{i_c}$ is the character of the irrep $i$ for the symmetry operation $c$, and $\chi_{r_c}$ is the character of the reducible representation $r$ for the symmetry operation $c$.
This allows us to write the irreps of the three $\ce{F}$'s orbitals.

$3\ce{F}$: $a_1'+a_2'+a_2''+2e'+e''$

With all the formalities out of the way, we can construct the qualitative MO diagram for $\ce{BF3}$. Like much of this post, my source for the MO diagram is the pdf of lecture notes for a Dartmouth chemistry course located at http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

Working our way up through this diagram, we see that we start by assuming that the SALCs of the $s$ orbitals of the three $\ce{F}$'s are assumed to be low enough in energy where they do not interact with the orbitals of $\ce{B}$.
The next point of interest is the $e' (\sigma)$ MO, which we form from the $p_x$ SALC of the $\ce{F}$ atoms and the $p_x$ and $p_y$ of the $\ce{B}$. We could additionally consider the interaction of the $e'$ $p_y$ SALC to form a $\pi$ bonding interaction, but we will assume that, with the $\sigma$ interaction already present, this $\pi$ bonding interaction will be weak and thus can be neglected.
This previous point leads us to have a set of nonbonding orbitals $a_2'+e'(y)+e''$. We then also have antibonding orbitals for each of the bonding orbitals below.
As a check on our work, we can compare this to the somewhat more quantitative diagram given in Principles of Inorganic Chemistry by Brian PFennig using the Student Editon of Spartan:

The only noticeable difference here is that the nonbonding MOs are now placed with respect to their computed energy rather than being lumped together. The classification of orbitals as bonding, nonbonding, and antibonding seems to fit with the diagram we have above, with the caveat that this more quantitative diagram does suggest some $e'(y)$ bonding contribution, as its energy is somewhat lower than the p orbitals it came from.
Now that we have the correct diagram, we can then compute the bond order for each $\ce{B-F}$ using: $$\frac{\text{Bonding electrons } -\text{Antibonding electrons}}{2\cdot\text{number of bonds}}$$ which for this diagram gives a value of $\frac{4}{3}$ per bond. This gives a sense of why $\ce{BF_3}$ is a strong Lewis acid, since it can accept a pair of electrons into its antibonding LUMO and still leave a compound with a BO of $1$ per bond.
As a final point, it is worth noting that this method of classifying orbitals is far from perfect and in general we should really look at individual bonds to classify if the orbital is anti/bonding with respect to that bond. However this energy based approach is what is commonly taught in school and happens to work well in this simple case; just be aware that it can fail for even seemingly simpler molecules like $\ce{CO}$.