How did he find the second derivative of $\theta$ with respect to $t$ ? $$\frac{\mathrm d^2 \theta}{\mathrm d t^2} = \omega^2 \frac{\mathrm d^2 \theta }{\mathrm d \tau^2}$$
$\tau = \omega t$ The first derivative $\tau'= \omega$, however the second derivative, $\omega$ is a constant independent of $t$. So the second derivative should be $0$ shouldnt it?
As usual, the problem appears partly to be a not clearly explained abuse of notation. The symbol $θ$ is used as coordinate, but implicitly also as symbol for two different functions. It might be clearer of one uses a longer formalism separating these roles with different symbols, using $θ=f(t)$ for the original function and $θ=g(τ)=g(ωt)$ for the function depending on the modified time scale. Then for the derivatives you get $$ \frac{dθ}{dt}=g'(τ)\frac{dτ}{dt}=g'(τ)ω\\ \frac{d^2θ}{dt^2}=g''(τ)\left(\frac{dτ}{dt}\right)^2+g'(τ)\frac{d^2τ}{dt^2}=g''(τ)ω^2\\ $$ The derivation prime denotes the derivative with respect to the function argument, whatever that is currently. And yes, $τ'(t)=$, $τ''(t)=0$, which is also used above.
Now one can also write in the $(τ,θ)$ dependency $g'(τ)=\frac{dθ}{dτ}$ and $g''(τ)=\frac{d^2θ}{dτ^2}$ to get back to the formulas that were originally in question.