UPDATED below.
In special relativity,
the two diagrams are consistent with each other.
To see this, I will draw in the tickmarks for space and time according to special relativity.
But first, some comments on your statement:
If the scaling is wrong, why is this picture so often used to explain length contraction? In images such as the one below, it seems to be assumed that the scale of both axes are the same.
- It is the Euclidean-scaling that is wrong (that is, inappropriate) for special relativity.
Instead, one should use the scaling according to Minkowski-spacetime geometry,
which [as you mention] uses the hyperbola as the "circle".
- In the image below (taken from Takeuchi's book) ,
it is not true that the "scale of both axes are the same". (The ratio of the drawn Euclidean-lengths is about 1.5 [using a ruler]. However, the ratio should be about 1.15 ($\approx 2/\sqrt{3}$, which is $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$ when evaluated for the velocity used in the diagram: $v=\frac{1}{2}c$.)
So, again, the Euclidean-scaling is inappropriate.
Let me draw in the tickmarks in accordance with special-relativity.
It turns out that using "rotated-graph paper" makes this easier to do,
where the grid is modeled after the spacetime-paths of light rays
for a light-clock carried by the lab-frame.
Instead of $(1/2)c$,
I'm going to use $v=(3/5)c$ because the arithmetic is easier is thus easier to draw in the grid. For my choice, $\gamma=5/4=1.25$ and $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=2$.
(side comment: Velocities $v$ with rational Doppler factors $k$ lead to Pythagorean Triples, which leads to calculations with fractions.)
Here your first diagram with the tickmarks included.
Note the symmetry of the situation: Each astronaut
measures the length of the other astronaut's ruler to be shorter by same factor $\gamma$.
![RRGP-robphy-lengthContraction-symmetric](https://i.stack.imgur.com/qFtqC.png)
For the diagram from Takeuchi's book, I have drawn in the
analogous features.
![RRGP-robphy-LengthContraction-Takeuchi](https://i.stack.imgur.com/PVL1pl.png)
(side comment:
The ticks from each clock ("light-clock-diamonds") are equal in area, in accordance with
the Lorentz boost transformation having determinant 1.
The grid directions in the rotated grid are along the light-cones of Minkowski spacetime [along the eigenvectors of the boost, with the Doppler factor and its reciprocal as their eigenvalues]. Hence, the boosted light-clock-diamond is "stretched" by a factor of $k=2$ is the future-frontward direction and by $1/k$ in the future-backward direction.)
Further info, see my answers in:
update:
This supports the comment I made in response to the OP's comment.
I made a Desmos visualization with Takeuchi's diagram and manually rescaled and positioned it as well as I could to match the grid.
Using the coordinates of the pairs-of-events each astronaut uses to measure the length of the moving rod, we approximately verify the length-contraction factor by calculating the ratio of computed-spatial-intervals using the $(t,x)$-coordinates of the lab frame.
Thus, the graph paper is accurate for describing the events in the lab frame.
However, this grid does not work for the moving frame. One cannot rotate this grid.... one must apply a lorentz-transformation to the grid. (My "rotated graph paper" makes this easier to do.)
https://www.desmos.com/calculator/cqvbvs8u9k
![robphy-Desmos](https://i.stack.imgur.com/4zDKY.png)
(Note: Takeuchi's example uses $v=(1/2)c$.)
Using the notation of my Desmos file,
$E=\left(7.5,7\right)$
$A=\left(3.5,7\right)$
$B=\left(2.165,4.35\right)$
$T_{diffLAB}=\left(E-A\right).y=0$
$X_{diffLAB}=\left(E-A\right).x=4$
$T_{diffAST}=\left(E-B\right).y=2.65$
$X_{diffAST}=\left(E-B\right).x=5.335$
$L_{ASTwrtAST}=\sqrt{X_{diffAST}^{2}-T_{diffAST}^{2}}=4.63030506554$
--this is not the Euclidean ruler-length on this diagram!
(Check it with a ruler and the Euclidean Pythagorean theorem!)
$L_{ASTwrtLAB}=\sqrt{X_{diffLAB}^{2}-T_{diffLAB}^{2}}=4$
$r_{GRAPHICAL}=\frac{L_{ASTwrtAST}}{L_{ASTwrtLAB}}=1.15757626639$
$g_{amma}=\frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^{2}}}=\frac{2}{\sqrt{3}}=1.15470053838 $
...reasonably good agreement.
update:
Working backwards, with knowledge of $v=(1/2)$ and $\gamma=\frac{2}{\sqrt{3}}$ and that observed-length is 4,
the coordinates of $B$ should be:
$B_x=7.5-\gamma(\gamma(4)+v\gamma(0))=2.1666$
$B_t=7-\gamma(\gamma(0)+v\gamma(4))=4.3333$
So, the bottom line is:
Takeuchi's grid is for lab-frame coordinates of events.
Determine the appropriate pairs of events for each astronaut's measurement.
Use the formula for the interval for each pair, and take the ratio.
Takeuchi's grid is not for moving-frame coordinates of events.
My "rotated graph paper" grid
helps visually compare the intervals quantitatively,
and can be more easily transformed to obtain the moving-frame
coordinates.
(You can still use the formulas, but my method can obtain the results graphically for velocities with rational Doppler factors.)
Best Answer
[While this may be unfamiliar...] the $ct'$ and $x'$ are Minkowski-perpendicular to each other in all frames.
In the reference frame of the $ct'$-axis, the axes will also appear to be [ordinary-Euclidean]-perpendicular to each other.
Here's how Minkowski describes this...
In other words,
locate the intersection of an observer's 4-velocity with the unit-hyperbola (the Minkowski circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity. That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.
The "intuition" to have is that
the tangent to the "circle" in that geometry
is
orthogonal to the radius vector.
You can play around with this idea in my visualization [screencaptured below].
This figure is unchanged by a Lorentz boost.
https://www.desmos.com/calculator/r4eij6f9vw
Play around with the E-slider to see the Galilean and Euclidean analogues!