Multiple Resonances along a transmission line
Exciting each resonance requires a different signal frequency.
Suppose the first resonance occurs at 50 MHz. Then the 2nd one will be at 100 MHz, the 3rd at 150 MHz, and so on.
If you excite the line with all these frequencies simultaneously, then you can produce a superposition of all the different standing wave patterns at the same time.
If you excite the line with only one frequency at a time, then you'll only excite one resonance at a time.

is it the same for the waveguide in the second picture? Or is it how I've always been told (&quot;Given an excitation frequency, there will be all resonance modes above cut-off?&quot;)?

The two situations are different. In the rectangular waveguide, if you express the modes as a combination of plane waves, each plane wave component has a momentum vector
$$\vec{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}, $$
and
$$\left|\vec{k}\right| = \frac{2\pi f}{c}$$
and the boundary conditions for each mode are
$$ k_x = \frac{n}{2}\frac{2\pi}{w_x}, n=1,2,3,..$$
$$ k_y = \frac{m}{2}\frac{2\pi}{w_y}, m=1,2,3,..$$
where \$w_x\$ and \$w_y\$ are the transverse dimensions of the waveguide.
Notice there is no restriction on \$k_z\$. That means for any \$f\$ above a given mode's cut-off we can find a \$k_z\$ that allows \$k_x\$ and \$k_y\$ to &quot;fit&quot; in the transverse dimensions and thus allows the mode to propagate.
In the longitudinal resonance case, you're already assuming a particular transverse mode, so \$k_x\$ and \$k_y\$ are fixed for any particular \$f\$ and there's no such additional degree of freedom available to allow \$k_z\$ (aka \$\beta\$) to vary to obtain a resonance except at the specific wavelengths where
$$\frac{2\pi f}{c}=k_z=\frac{n}{2}\frac{2\pi}{\ell}$$