It's hard to do a one-to-one comparison. For starters they assume a spring a zero free length for "simplicity" ( I find that somewhat amusing given the analysis that follows). I understand their motivation for that, they only care how it moves, not where it moves. Anyhow, seeing that assumption helped me to question what I was messing up in (1). ( the free length of the spring )
Let ## l_o ## be the free length of the spring, and ## s(t) ## be the required deflection for equilibrium:
$$ x_o(t) = l_o + s(t) $$
From this (1) becomes:
$$ \begin{align} M \frac{d^2}{dt^2} \left( x - x_o(t) \right) &= -k \left( s + ( x - ( l_o + s ) ) \right) + Mg \tag*{} \\ M \frac{d^2}{dt^2} \left( x - ( l_o + s) \right) &= -k \left( x - l_o \right) + Mg \tag*{} \\ M \frac{d^2}{dt^2} \left( x - s \right) &= -k \left( x - l_o \right) + Mg \tag*{} \end{align} $$
With
$$ s = \frac{Mg}{k} $$
It follows that (1) is given by:
$$ M \ddot x + k x = \left( 1 + \frac{ \ddot M}{k} \right) M g + k l_o \tag{1} $$
As a verification if we plug in that the hole is plugged ##\dot M = 0## and ## x = x_o ## we find that:
$$ M \ddot x + k \left( l_o + s\right) = Mg + k l_o $$
$$ M \ddot x = Mg - ks $$
but ## Mg - ks \equiv 0 ##
So as we expect, the system remains static, i.e. ## M \ddot x = 0 ##
I'm more settled on that now.
Also, the authors in the Princeton paper are including the "thrust force" from the jet.
Furthermore, they do a correction for Bernoulli's in an accelerated frame, but it's a little complex for me to tell what is happening there in comparison to how I "tried" to correct for it.