Derivation of the acceleration of an observer through direct use of the Lorentz transformation

There are actually several ways one can derive the expression for the coordinates moving with constant acceleration, if you consider a bit more directly the geometry of curves in Minkowski space, but that's a deviation from your question. The formula you are interested in can be derived directly from the chain rule in calculus. It is just that the derivation is tedious.

I am going to start with the most general version of this formula. AFter that will reduce the result to the case of Lorenz transformations.

Assume you have a time-line curve $big(, t ,, , x(t) ,big)$ and at least twice continuously differentiable change of coordinates (curvilinear in general): begin{align} &x' = x'(x, t)\ &t' = t'(x, t) end{align} your curve becomes a curve that can be parametrized as $big(, t' ,, , x'(t') ,big)$ where begin{align} &x' = x'big(x(t),, tbig)\ &t' = t'big(x(t),, tbig) end{align} For example, in the case of Lorenz transformations begin{align} &x' = gamma ,big(x , - , V,tbig)\ &t' =gamma ,big(-, V, x , + , tbig) &text{ where } , gamma = frac{1}{sqrt{1-V^2}} end{align} By the chain rule begin{align} frac{dx'}{dt'} , &= , frac{dt}{dt'}, frac{dx'}{dt} , = , frac{1}{left( frac{dt'}{dt}right)} , frac{dx'}{dt} end{align} Let us calculate both factors begin{align} &frac{dx'}{dt} , = , frac{partial x'}{partial x}, frac{dx}{dt} , + , frac{partial x'}{partial t}\ &frac{dt'}{dt} , = , frac{partial t'}{partial x}, frac{dx}{dt} , + , frac{partial t'}{partial t} end{align} Thus, the first derivative becomes begin{align} frac{dx'}{dt'} , = , frac{,frac{partial x'}{partial x}, frac{dx}{dt} , + , frac{partial x',}{partial t}}{ frac{partial t'}{partial x}, frac{dx}{dt} , + , frac{partial t'}{partial t}} end{align} For example, in the case of Lorenz, begin{align} &frac{dx'}{dt} , =, gamma ,left(frac{dx}{dt} , - , Vright)\ &frac{dt'}{dt} ,=, gamma ,left(-, V, frac{dx}{dt} , + , 1right) &text{ where } , gamma = frac{1}{sqrt{1-V^2}} end{align} $$ frac{dx'}{dt'} , = , frac{, frac{dx}{dt} , - , V}{1, -, V, frac{dx}{dt} ,} , = , frac{, frac{dx}{dt} , - , V ,}{1 , - , V, frac{dx}{dt}} $$ Now, for the second derivative begin{align} frac{d^2x'}{dt'^2} , =& , frac{d}{dt'}left( frac{dx'}{dt'}right) , = , frac{d}{dt'}left( frac{1}{left(frac{dt'}{dt}right)}, frac{dx'}{dt} right) \ =& , frac{dt}{dt'},frac{d}{dt}left( frac{1}{left(frac{dt'}{dt}right)}, frac{dx'}{dt} right) , = , frac{dt}{dt'},left[, frac{d}{dt}left( left(frac{dt'}{dt}right)^{-1} ,right) , frac{dx'}{dt} , + , left(frac{dt'}{dt}right)^{-1} , frac{d^2x'}{dt^2} , right] \ =& , frac{dt}{dt'},left[, - , left(frac{dt'}{dt}right)^{-2} , frac{dx'}{dt} , frac{d^2t'}{dt^2}, + , left(frac{dt'}{dt}right)^{-1} , frac{d^2x'}{dt^2} , right] \ =& , left(frac{dt'}{dt}right)^{-1},left[, - , left(frac{dt'}{dt}right)^{-2} , frac{dx'}{dt} , frac{d^2t'}{dt^2}, + , left(frac{dt'}{dt}right)^{-1} , frac{d^2x'}{dt^2} , right] \ =& , - , left(frac{dt'}{dt}right)^{-3} , frac{dx'}{dt} , frac{d^2t'}{dt^2} , + , left(frac{dt'}{dt}right)^{-2} , frac{d^2x'}{dt^2} \ =& , left(frac{dt'}{dt}right)^{-2} , frac{d^2x'}{dt^2} , - , left(frac{dt'}{dt}right)^{-3} , frac{dx'}{dt} , frac{d^2t'}{dt^2}\ =& , left(frac{dt'}{dt}right)^{-3} ,left[, frac{dt'}{dt},frac{d^2x'}{dt^2} , - , frac{dx'}{dt} , frac{d^2t'}{dt^2} ,right]\ end{align} where the formulas for the second derivatives can be derived fro mthe chain rule as follows begin{align} frac{d^2x'}{dt^2} , =& , frac{d}{dt} left(frac{dx'}{dt}right) , = , frac{d}{dt} left(frac{partial x'}{partial x}, frac{dx}{dt} , + , frac{partial x'}{partial t}right) \ =& , frac{partial x'}{partial x}, frac{d^2x}{dt^2} , + , frac{partial^2 x'}{partial x^2},left(frac{dx}{dt}right)^2, + , 2, frac{partial^2 x'}{partial x partial t}, frac{dx}{dt} , + , frac{partial^2 x'}{partial t^2} \ &\ frac{d^2t'}{dt^2} , =& , frac{d}{dt} left(frac{dt'}{dt}right) , = , frac{d}{dt} left(frac{partial t'}{partial x}, frac{dx}{dt} , + , frac{partial t'}{partial t}right) \ =& , frac{partial t'}{partial x}, frac{d^2x}{dt^2} , + , frac{partial^2 t'}{partial x^2},left(frac{dx}{dt}right)^2, + , 2, frac{partial^2 t'}{partial x partial t}, frac{dx}{dt} , + , frac{partial^2 t'}{partial t^2} end{align} So to summarize, $$frac{d^2x'}{dt'^2}, = , left(frac{dt'}{dt}right)^{-3} ,left[, frac{dt'}{dt},frac{d^2x'}{dt^2} , - , frac{dx'}{dt} , frac{d^2t'}{dt^2} ,right]\$$ where begin{align} &frac{dx'}{dt} , = , frac{partial x'}{partial x}, frac{dx}{dt} , + , frac{partial x'}{partial t}\ &frac{dt'}{dt} , = , frac{partial t'}{partial x}, frac{dx}{dt} , + , frac{partial t'}{partial t}\ frac{d^2x'}{dt^2} , =& ,frac{partial x'}{partial x}, frac{d^2x}{dt^2} , + , frac{partial^2 x'}{partial x^2},left(frac{dx}{dt}right)^2, + , 2, frac{partial^2 x'}{partial x partial t}, frac{dx}{dt} , + , frac{partial^2 x'}{partial t^2} \ frac{d^2t'}{dt^2}, =& , frac{partial t'}{partial x}, frac{d^2x}{dt^2} , + , frac{partial^2 t'}{partial x^2},left(frac{dx}{dt}right)^2, + , 2, frac{partial^2 t'}{partial x partial t}, frac{dx}{dt} , + , frac{partial^2 t'}{partial t^2} end{align} In the case of Lorenz transformations, begin{align} &frac{dx'}{dt} , =, gamma ,left(frac{dx}{dt} , - , Vright)\ &frac{dt'}{dt} ,=, gamma ,left(-, V, frac{dx}{dt} , + , 1right)\ &frac{d^2x'}{dt^2} , =, gamma ,frac{d^2x}{dt^2} \ &frac{d^2t'}{dt^2} ,=, -,gamma, V, frac{d^2x}{dt^2}\ &text{ where } , gamma = frac{1}{sqrt{1-V^2}} end{align} Thus $$frac{d^2x'}{dt'^2}, = , left(,-,gamma, V, frac{dx}{dt} , + , gamma,right)^{-3} ,left[, left(-, gamma, V, frac{dx}{dt} , + , gammaright) , gamma ,frac{d^2x}{dt^2} , + , left(gamma,frac{dx}{dt} , - , gamma,Vright), gamma, V, frac{d^2x}{dt^2} ,right]$$ $$= , left(,-,gamma, V, frac{dx}{dt} , + , gamma,right)^{-3} ,left[, left(1, -, V, frac{dx}{dt} , right) , + , left(frac{dx}{dt} , - ,Vright), V, right] , gamma^2 ,frac{d^2x}{dt^2} $$ $$= , gamma^{-3},left(,1, - ,V, frac{dx}{dt} , right)^{-3} ,left[, 1, -, V, frac{dx}{dt} , + , V,frac{dx}{dt} , - ,V^2, right] , gamma^2 ,frac{d^2x}{dt^2}$$ $$= , gamma^{-1},left(,1, - ,V, frac{dx}{dt} , right)^{-3} ,big(, 1, - ,V^2, big) ,frac{d^2x}{dt^2}$$ $$= , frac{,gamma^{-1} ,big(, 1, - ,V^2, big) ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$ $$= , frac{,big(, 1, - ,V^2, big)^{frac{1}{2}} ,big(, 1, - ,V^2, big) ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$ And here is the formula

$$frac{d^2x'}{dt'^2}, = , frac{,big(, 1, - ,V^2, big)^{frac{3}{2}} ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$

What have you tried?
It might help to see your attempt to see where you got stuck.

Try using the chain rule:

$$ begin{align} color{red}{frac{dx'}{dt'}} =frac{dx'}{dt}frac{dt}{dt'} &=frac{left(displaystylefrac{dx'}{dt}right)}{left(displaystylefrac{dt'}{dt}right)}\ &=frac{ displaystylefrac{d}{dt}left( frac{x-Vt}{(1-V^2)^{1/2}} right) }{displaystyle frac{d}{dt}left( frac{t-Vx}{(1-V^2)^{1/2}} right) } end{align} $$ Then, $$ begin{align} frac{d^2 x'}{dt'^2}=frac{d}{dt'}left( color{red}{frac{dx'}{dt'}} right) =frac{dleft( {frac{dx'}{dt'}} right)}{dt}frac{dt}{dt'} &=frac{displaystyleleft(frac{dleft( frac{dx'}{dt'} right)}{dt}right)}{left(displaystylefrac{dt'}{dt}right)} \ &=frac{displaystylefrac{d}{dt}left( color{red}{frac{dx'}{dt'}}right)} {displaystyle frac{d}{dt}left( frac{t-Vx}{(1-V^2)^{1/2}} right)} end{align} $$