$\lambda_{max}(A) \leq \lambda_{min}(B)$ is a sufficient property for the so-called "Loewner order".
See (https://en.wikipedia.org/wiki/Loewner_order)
But is is not a necessary condition. Here is a counterexample; take:
$$A=\pmatrix{1&1\\1&2} \ \ \text{and} \ \ B=\pmatrix{2&2\\2&3}.$$
We have: $A \leq B$ for the Loewner order because $B-A=\pmatrix{1&1\\1&1}$ is semi-positive-definite (with eigenvalues $0$ and $2$).
The spectra of $A$ and $B$ are resp.
$$\{0.3819,2.6180\} \ \ \text{and} \ \ \{0.4384, 4.5616\}.$$
but without having $\lambda_{max}(A) \leq \lambda_{min}(B).$
Yeah, there seems to be a little bit of confusion here, but you seem to have the right idea. This might just boil down to a typo, that the author meant to say eigenvalue but instead typed singular value. You should ask the author about it.
For a symmetric matrix $A$, the singular values are the absolute values of the eigenvalues. So, as you point out, if $A$ is positive semidefinite, they are the same.
To point out where the result stated in the link is wrong, consider:
$$A = \begin{pmatrix} -2 & 0 \\ 0 & 1 \end{pmatrix}$$
which has eigenvalues $-2, 1$, is symmetric and therefore has singular values $2, 1$. The quadratic form $x^TAx$ takes the form:
$$\begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} -2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =
\begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} -2x_1 \\ x_2 \end{pmatrix}
= -2x_1^2 + x_2^2
$$
which for $x_1^2 + x_2^2 = 1$ takes its maximum value $1$ at $(x_1, x_2) = (0,1)$. So, the function never takes the value $2$, the value of the largest singular value.
Note however, that if you would compute $|x^TAx|$, the absolute value of the quadratic form, the maximum would be the largest singular value.
In your second link, note that the formulation is slightly different: Here we have two vectors $x,y$ as input, and we compute $y^TAx$. The value of the largest singular value can be achieved by using the singular value's left- and right-singular vectors.
For the example above, you can take
$$y = \begin{pmatrix} -1 \\ 0 \end{pmatrix}
\quad
x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
to achieve the value of $2$ for $y^T A x$.
Best Answer
I suppose that you mean $\rho(A)<\rho(B)$, otherwise your question does not make sense, as the spectra of $A$ and $B$ may contain non-real eigenvalues. If so, the inequality is true and Rayleigh quotient has nothing to do with it. The inequality is a simple consequence of Perron-Frobenius theorem, as discussed here. In general, if $A$ and $B$ are two different nonnegative matrices such that $A+B$ is irreducible, then $\rho(A)<\rho(A+B)$.