Heuristically, an expander $G$ looks locally like the $d$-regular tree $T$. Let $r$ be a positive real number and let $f_x$ be the function on the vertices of $T$ given by $f_x(y) = r^{d(x,y)}$. We have $|f_x|^2 = 1+d \sum_{n=1}^{\infty} (d-1)^n r^{2n}$; this sum is convergent as long as $r < 1/\sqrt{d-1}$. Writing $A$ for the adjacency matrix, we have
$$A f_x = (r^{-1}+(d-1) r) f_x + (r-r^{-1}) \delta_x$$
where $\delta_x$ is the $\delta$ function at $x$. As $r \to \tfrac{1^{-}}{\sqrt{d-1}}$, the norm of $f_x$ grows while $(r-r^{-1}) \delta_x$ remains bounded. So, intuitively, at $r=1/\sqrt{d-1}$, the function $f$ acts like an eigenvector with eigenvalue $\sqrt{d-1} + (d-1)/\sqrt{d-1} = 2 \sqrt{d-1}$.

In an expander, $G$ still locally looks like $T$. Take two vertices $x$ and $y$ that are very far apart and consider $f_x - f_y$. This is orthogonal to $1$, so $\tfrac{\langle f_x-f_y, \ A (f_x - f_y) \rangle}{\langle f_x-f_y, \ f_x-f_y \rangle}$ is a lower bound for $\lambda_2$. Assuming we take $r < 1/\sqrt{d-1}$, the overlap between $f_x$ and $f_y$ will be negligible and the analysis will be dominated by the behavior near $x$ and $y$, which will look like the case of $T$.

For a rigorous proof of Alon-Boppana along vaguely these lines, see Theorem 5 here. I'm not claiming the arguments are very close, but they start with two points that are far apart and use them to build a function $g$ with $\langle g,\ 1 \rangle=0$ and $\tfrac{\langle g, A^{2k} g \rangle}{\langle g,g \rangle}$ large by using that $G$ locally looks like $T$.