What are the $sp^3$ orbital bond directions relative to a Cartesian reference frame? When I try to search for this, I find images like below, but it is not clear to me what the actual bond directions are, i.e. the first one appears to be (0, 0, 1) but then I am not sure of the others.

 A: Yes, the end-point of the first orbital is placed at $(0,0,1)$.
The end-points of the other three orbitals are placed in a such a way, so that


*

*each of them is at distance $1$ away from the origin $(0,0,0)$,

*and all four are separated from each other by the same distance.


By doing so you arrive at a configuration as described at
Chemistry glossary - sp3 hybrid orbital:

The four $sp^3$ hybrid orbitals point toward the corners
   of a regular tetrahedron with the bond angle of 109.5°.

A: 
When I try to search for this, I find images like below, but it is not
  clear to me what the actual bond directions are [...]

In order to minimise Coulombic repulsion between the $\text{sp}^3$ lobes, the bonds form a regular tetrahedron with the nucleus at the centre of it (as shown in the bottom picture). The inter-bond angles are thus maximised.
So the bonds do not lie on the main Cartesian axis.
A: You can convert the spherical coordinates to cartesian coordinates easily.  As you know,
$$\begin{align}
z&=r \cos(\phi) \\
y&=r \sin(\theta)\sin(\phi) \\
x&=r \cos(\theta)\sin(\phi)
\end{align}$$
Following your lead, you can choose $(0,0,1)$ [$z$ axis] as one of your unit vectors.  If you look from above, the $\theta$ values are $0$ and $120°$ in either direction.  The $\phi$ values are $109.5°$.  Choosing one of your unit vectors to be directly under the $x$ axis [$\theta=0$], the other three vectors, in spherical coordinates, are:
$$\begin{bmatrix}
           1 \\
           0° \\
           109.5° \\
         \end{bmatrix},
\begin{bmatrix}
           1 \\ 120° \\ 109.5° \\
         \end{bmatrix}, 
\begin{bmatrix}
           1 \\
           -120° \\
           109.5° \\
         \end{bmatrix}$$
Here is a handy online calculator to complete the calculation.
https://keisan.casio.com/exec/system/1359534351
It gives:
$$
\begin{bmatrix}0\\0\\1 \end{bmatrix},
\begin{bmatrix}0.942\\0\\-0.3338 \end{bmatrix},
\begin{bmatrix}-0.47\\0.816\\-0.338 \end{bmatrix},
\begin{bmatrix}-0.47\\-0.816\\-0.338 \end{bmatrix}
$$
