Why aren't the lengths of the bars on a toy glockenspiel proportional to the wavelengths?
The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ This case is the other way around: the tension is negligible, so we only have the 'extra' term. The wave equation becomes $$-A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ Plugging in an ansatz of $\cos(kx-\omega t)$ gives the dispersion relation $$Ak^4 = \omega^2.$$ That is, $\omega \propto k^2$. Since $k$ is inversely proportional to length, $$\omega \propto 1/L^2$$ as desired. A bar $\sqrt{2}$ times shorter makes a tone twice as high.
As you saw, the wave speed must change for the results to make sense. The phase velocity of a wave is $v_p = \omega / k$, and this is constant only for the simplest dispersion relation, the ideal wave equation $\omega = vk$. In this case, we have $\omega \propto k^2$, which implies $v_p \propto k$. Waves with shorter wavelength, like the ones on the smaller bars, travel faster.
But this doesn't mean anything about the smaller bars is different. The phase velocity changes because wave propagation is fundamentally different on bars than strings; it exhibits dispersion.
As knzhou identifies, the key difference between vibrations of a free beam and a string is that the restoring force is now provided by bending moments (proportional to $\frac{d^4y}{dx^4}$) rather than linear tension (proportional to $\frac{d^2y}{dx^2}$).
The consequence, as this source shows, is that for free beams like the bars of the glockenspiel, angular frequency $\omega=2\pi f$ and wavenumber $k=\frac{2\pi}{\lambda}$ are related by
$$\omega^2=\frac{YI}{\rho A} k^4$$
where $I=\frac{1}{12}bh^3$ is 2nd moment of cross-sectional area about a horizontal axis through the centre, and $A=hb$ is the cross-sectional area.
Because the bars are unconstrained at any point, the wavelength does not correspond exactly to a multiple of the bar length. However, in each mode the same relation applies for different bar lengths. In the fundamental mode the bars vibrate with nodes at approx. $0.224L$ from each end (source), so that $\lambda=1.104L$. This leads (I think) to
$$f=1.488\frac{h}{L^2}\sqrt{\frac{Y}{\rho}}$$
which is the same form as the formula quoted here, but the leading factor does not quite agree.
Frequency is inversely proportional to square of length, so halving the frequency requires increasing the length by a factor of $\sqrt2=1.414$ as you found.