I think the essence of your question is really just that the wonderful answer by @AndrewSteane did not include the trivial additions of mathematical expressions of what he was trying to say. My answer is thus of zero physical insight, but just paraphrasing him in mathematical terms.
4-velocity
The 4-velocity $\boldsymbol U$ corresponding to a 3-velocity $\vec v$ can be expressed by
$$\tag1\boldsymbol U=\frac{\mathrm d\boldsymbol x}{\mathrm d\tau}=
\begin{pmatrix}\cosh|\vec v|\\\hat{\vec v}\sinh|\vec v|
\end {pmatrix}
$$
and this construction clearly obeys $\boldsymbol U\cdot\boldsymbol U=1$. In this, I am not claiming that $\boldsymbol x$ is well-defined, only simply trying to connect to how people usually think/write things.
4-acceleration
The 4-acceleration is then defined as
$$\tag2\boldsymbol A=\frac{\mathrm d\ }{\mathrm d\tau}\boldsymbol U=
\begin{pmatrix}\sinh|\vec v|\\\hat{\vec v}\cosh|\vec v|
\end {pmatrix}\frac{\mathrm d|\vec v|}{\mathrm d\tau}+
\begin{pmatrix}0\\\frac{\mathrm d\ }{\mathrm d\tau}\hat{\vec v}
\end {pmatrix}\sinh|\vec v|
$$
where it is clear that both terms are orthogonal to $\boldsymbol U$ so that $\boldsymbol U\cdot\boldsymbol A=0$ as required.
For the rest of this discussion, we specialise to the case that $\frac{\mathrm d\ }{\mathrm d\tau}\hat{\vec v}=\vec0$ so that the particle is always only moving in the same 3-vector direction $\hat{\vec v}$ in a straight line. When this is the case, then we have $\boldsymbol A\cdot\boldsymbol A=-\left(\dfrac{\mathrm d|\vec v|}{\mathrm d\tau}\right)^2$
4-momentum and 4-force
The 4-momentum $\boldsymbol P=m_0\boldsymbol U$ and the 4-force $\boldsymbol F=\frac{\mathrm d\ }{\mathrm d\tau}\boldsymbol P=m_0\boldsymbol A$, but I would like to stress that the components of the 4-momentum is only ever
$$\tag3\boldsymbol P=
\begin{pmatrix}E/c\\\vec p
\end {pmatrix}
$$
and the answerers on the other question are all wrong when they assert that mass is changing. They are using the outdated concept of relativistic mass and that should never have been defined at all. The only well-defined concept of mass is rest mass $m_0$, and no other concept of mass should exist. We will come back to this in much greater detail.
Constant 4-force = Constant 4-acceleration, but is unphysical
It should be clear from the above definitions and discussions that if $\boldsymbol F=(F_0,\vec F)^T$ is constant, then so would the 4-acceleration, and then by the momentum derivative definition of 4-force, that means
$$\tag4\boldsymbol P=
\begin{pmatrix}E_i/c+\tau F_0\\\vec p_i+\tau\vec F
\end {pmatrix}
$$
It is then interesting to try and compute its invariant rest energy = rest mass$/c^2$.
$$
\begin{align}\left(E_i+\tau F_0\right)^2-(\vec p_i+\tau \vec F)^2
\tag5&=(E_i^2-\vec p_i^2)+\tau^2(F_0^2-\vec F^2)+2\tau(E_iF_0-\vec p_i\cdot\vec F)\\
\tag6&=m_0^2-m_0^2\tau^2\left(\frac{\mathrm d|\vec v|}{\mathrm d\tau}\right)^2
+2\tau(E_iF_0-\vec p_i\cdot\vec F)
\end {align}
$$
specialising to the initially at-rest case where $\vec p_i=\vec 0$ and $E_i=m_0$, this means that the initial $|\vec v|=0$, which zeroes out the $\sinh$ function, so that $F_0=0$, we see that the invariant rest energy of this particle would obey
$$\tag7m=m_0\sqrt{1-\left(\tau\frac{\mathrm d|\vec v|}{\mathrm d\tau}\right)^2}\qquad\text{which is extremely unphysical}$$
Unphysical because it means that this particle would be "losing rest mass" just by executing such an acceleration. Let us not mince words: such a scenario is physically impossible to achieve. An examination body should be fined by the government if it appears in one of their examinations.
Constant 3-velocity direction but with constant 4-force magnitude = constant 4-acceleration magnitude
Looking at Equation (1), it is clear that even though a particle may be moving in a straight line, in one fixed 3-velocity direction $\hat{\vec v}$, if it is accelerating, then it would be "rotating" in the "Energy-Momentum Minkowski diagram", because $|\vec v|$ is increasing and thus the $\cosh$ and the $\sinh$ will be changing, doing this hyperbolic rotation.
It should be trivial to see that, by reversing the definitions all the quantities involved, that
$$\tag8\boldsymbol P_f=\boldsymbol P_i+\int\boldsymbol F\,\mathrm dt=\boldsymbol P_i+m_0
\begin{pmatrix}\cosh|\vec v|_f-\cosh|\vec v|_i\\\hat{\vec v}(\sinh|\vec v|_f-\sinh|\vec v|_i)
\end {pmatrix}
$$
With the pleasing consequence that the invariant rest mass of the particle is always the same. No weirdness. This is the physically allowed scenario that some textbooks cover in greater detail by working out the positions as a function of proper time and whatnot.
The unphysicality of constant 4-force and the physicality of this, is what the answers are trying to say.
The case of constant 3-force does not seem interesting to me and so I am not going to even treat it.
