Two functions $g(n)$, $G(n)$ such that $g(n) = o(G(n))$ but $g(n+1) \neq o(G(n))$
One example is $g(n) = 2^{2^{n-1}}$, $G(n) =2^{2^{n}}$.

We have $g(n) = \sqrt{G(n)}$, so $g(n) = o(G(n))$.

Meanwhile, of course, $g(n+1) = G(n)$ so $g(n+1) = \Theta(G(n))$, so $g(n+1)\neq o(G(n))$

Why double exponential?  When we add one to $n$, we want the effect to be more than a multiplicative constant (because big $O$ notation hides multiplicative, and therefore additive constants).  Let's make it simple and say that we want adding 1 to $n$ to have a polynomial effect on the value of $g(n)$.  When you add a constant to $n$:


a linear function is increased by an additive constant
an exponential function is increased by a multiplicative constant
a double exponential function is increased polynomially (by a power of two in our example, hence $g(n)^2 = G(n)$)


We can also have $g(n) = n^n$, $g(n) = n!$, etc., where $G(n) = g(n-1)$.  In general we can let $g(n) = f(n)^n$, where $f(n) = \omega(1)$ and $f$ is nondecreasing.  Then we have:

$$\lim_{n\to\infty} \frac{g(n)}{G(n)} = \lim_{n\to\infty} \frac{g(n)}{g(n+1)} = \lim_{n\to\infty}\frac{f(n)^n}{f(n+1)^{n+1}} \leq \lim_{n\to\infty}\frac{f(n)^n}{f(n)^{n+1}} = \lim_{n\to\infty}\frac{1}{f(n)} = 0$$

So we've shown $g(n) = o(G(n))$ using the limit definition (the last step is because $f(n) = \omega(1)$).  Therefore, functions like $(\log\log n)^n$ and even $\alpha(n)^n$ where $\alpha$ is the inverse Ackermann function, will do the trick for us as well.