Analyzing Example 13.6 and Example 13.7
If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3
Except Strauss in his usual manner to tell truth, only truth, but never all the truth nixes the case of 2 negative eigenvalues as he assumes that either $\alpha>0$ or $\beta>0$. The easiest way to deal with it would be to note that $\alpha+\beta+\alpha\beta l=0$ has two branches and divides plane into 3 regions and due to continuity of e.v. in each of them the number of negative eigenvalues is the same. Consider $\beta=\alpha$, it transects all three regions. Shift coordinate $x$ to the center of interval, which becomes $[-L,L]$, $L=l/2$. Now problem becomes \begin{align} &X''+\lambda X=0,\label{eq-1}\\ &X'(-L)=\alpha X(-L),\label{eq-2}\\ &X'(L)=-\alpha X(L)\label{eq-3} \end{align} and therefore if $X$ is an eigenfunction, then $Y(x):=X(-x)$ is e.f. with the same e.v.
Therefore we can consider separately e.f. which are even functions, and which are odd functions--and those are described respectively by \begin{align} &X''+\lambda X=0,\label{eq-4}\\ &X'(0)=0,\label{eq-5}\\ &X'(L)=-\alpha X(L)\label{eq-6} \end{align} and \begin{align} &X''+\lambda X=0,\label{eq-7}\\ &X(0)=0,\label{eq-8}\\ &X'(L)=-\alpha X(L).\label{eq-9} \end{align} Since we are looking at $\lambda=-\gamma^2$ ($\gamma>0$, we look at $X=\cosh (x \gamma)$ and $X=\sinh (X\gamma)$ respectively (see conditions (\ref{eq-5}), (\ref{eq-8}) and then conditions (\ref{eq-6}), (\ref{eq-9}) tell us that \begin{align} &\alpha L = -(L\gamma)\tanh (L\gamma),\label{eq-10}\\ &\alpha L= - (L\gamma) \coth (L\gamma)\label{eq-11} \end{align} respectively.
Both functions $z\tanh(z)$ and $z\coth(z)$ are monotone increasing for $z>0$ with minima at $z=0$ equal $0$ and $1$ respectively. Thus equation (\ref{eq-10}) has a single solution $\gamma$ iff $\alpha<0$ and (\ref{eq-11}) has a single solution $\gamma$ iff $\alpha L < -1$.
Therefore as $\alpha l<0$ there is one negative eigenvalue with even e.f. and as $2\alpha l+(\alpha l)^2<0$ comes another negative eigenvalue with odd e.f.
Sure, one can apply a variational argument: but analysis above has its own merit (mainly learning).