If you take the functional derivative of first order, i.e $\frac{\delta S[x(t)]}{\delta x(t)}=- \ddot{x}(t)+V'(x)$.

Then express these with dirac deltas i.e:

$\int \ddot{x}(s)\delta(s-t')ds= \int x(s)\ddot{\delta}(s-t')ds$ where I have used to integration by parts and boundary conditions obviously vanish.

So if you take another functional derivative you get:

$$\int \frac{\delta x(s)}{\delta x(t)} \ddot{ \delta}(s-t') ds = \int \delta(t-s) \ddot{\delta}(s-t') ds = \ddot{\delta}(t-t')$$

And for the other part also:

$$V'(x(t')) = \int V'(x(s)) \delta(s-t')ds$$

So now we have:

$$\int \frac{\delta V'(x(s))}{\delta x(t)} \delta(s-t')ds = \int V''(x(s)) \delta(s-t)\delta(s-t') ds = V''(x(t))\delta(t-t')$$

So if we gather all the terms we get:

$$-\ddot{\delta}(t-t')+V''(x(t))\delta(t-t')$$

Which is a rigorous treatment of the maple calculation.