1) No, you can't assume that the eigenfunctions are even or odd. You can only do that when the Hamiltonian commutes with the parity operator.
2) I guess you could try deducing something from that, e.g. using perturbation theory with $\delta_{-r}$ as the perturbation, but I wouldn't go down that road; it doesn't seem very promising.
3) I don't know whether he "saw" that the operator has exactly two negative eigenvalues – I presume he says that because he performed the calculation. Regarding the boundary conditions: The jump discontinuity is in the first derivative, not the second; the second derivative has a delta peak at that point, since it has to cancel the delta peak from the potential in the Schrödinger equation. Integrating the second derivative yields the first derivative, and integrating over $-2\delta f$ yields a jump of height $-2f$; since there's no jump in $f$ itself, these two have to be equal, so the first derivative must have a jump of height $-2f$.
[Edit in response to the comment:]
Away from the delta peaks, for negative eigenvalues the solution is a superposition of two exponentials decaying towards positive and negative $x$ values, respectively. There are three regions, left, right and centre. In the left region there can be no leftward increasing component, and in the right region there can be no rightward increasing component. That leaves four unknown amplitudes, of which we can arbitrarily set one to $1$ since the wavefunction will be normalized:
$ f(x)= \begin{cases} \mathrm e^{\lambda x}&x \le -r\;,\\ b_+\mathrm e^{\lambda x}+b_-\mathrm e^{-\lambda x}&-r \lt x \le r\;,\\ c\mathrm e^{-\lambda x}&r \lt x\;.\\ \end{cases} $
As you wrote, in the symmetric case, we can assume that the wavefunction has definite parity. For positive parity, we get
$ f_+(x)= \begin{cases} \mathrm e^{\lambda x}&x \le -r\;,\\ b\mathrm e^{\lambda x}+b\mathrm e^{-\lambda x}&-r \lt x \le r\;,\\ \mathrm e^{-\lambda x}&r \lt x\;.\\ \end{cases} $
The continuity condition is
$ \mathrm e^{-\lambda r} = b\mathrm e^{\lambda r}+b\mathrm e^{-\lambda r}\;, $
and the jump condition that you wrote is
$ -\lambda\mathrm e^{-\lambda r}-\left(\lambda b\mathrm e^{\lambda r}-\lambda b\mathrm e^{-\lambda r}\right)=-2\mathrm e^{-\lambda r}\;. $
(There's only one of each now because of the symmetry.)
We can solve the first condition for $b$ and substitute it into the second:
$b=\frac{\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}}\;,$
$ -\lambda\mathrm e^{-\lambda r}-\lambda\frac{\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}}\left(\mathrm e^{\lambda r}-\mathrm e^{-\lambda r}\right)=-2\mathrm e^{-\lambda r}\;. $
Dividing by $-\lambda\mathrm e^{-\lambda r}$ yields
$ \begin{align} 1+\frac{\mathrm e^{\lambda r}-\mathrm e^{-\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}} &= \frac2\lambda\;,\\\\ \frac{\mathrm e^{\lambda r}}{\mathrm e^{\lambda r}+\mathrm e^{-\lambda r}} &= \frac1\lambda\;,\\\\ 1+\mathrm e^{-2\lambda r}&=\lambda\;. \end{align} $
Since the left-hand side is strictly decreasing and the right-hand side is strictly increasing, this equation has exactly one solution, which for $r=1$ Wolfram|Alpha locates at $\lambda \approx1.10886$.
You can do the same thing for negative parity to find the second eigenvalue. In the asymmetric case, you'll have to do a little more work, since you have to keep all three constants if you can't use symmetry to simplify.
P.S.: If you're wondering how come $\lambda$ occurs without $r$ in the equation even though $r$ seems to be the only length scale in the problem: There's a hidden length scale because the jump in $f'$ should have units of inverse length, so the delta strength $2$ introduces a characteristic length $1/2$.