For questions like this, it can help to first consider the CDF. We can then take the derivative and have our density. I'll give you some hints and you need to fill in the gaps.
Ask yourself: What is the CDF of $Y_2^N$, the second highest valuation of $N$ iid draws? It is the probability that the realization of random variable $Y_2^N$ is less than some $x$. Next, ask yourself two auxiliary questions: What is the probability that the highest value is less than $x$? (It's $F^N(x)$) What is the probability that exactly one of $N$ values is higher than $x$ and $(N-1)$ values are lower than $x$? (It's $N (1-F(x))F^{N-1}(x)$ as there are $N$ candidates to be the one above $x$!) Once you have answered these two questions, you can think about the first question again: What is the probability that the second-highest value is at most $x$? Exactly, the CDF is the sum of the answer to the earlier questions:
$$F_2^N (x) = (F(x))^N + N (1-F(x)) (F(x))^{N-1}= N F^{N-1}(x) - (N-1)F^N(x),$$
and the derivative is
$$f_2^N(x) = N(N-1) f(x) (F^{N-2}(x) - F^{N-1}(x)) = N(N-1)f(x) F^{N-2}(x)(1-F(x)).$$
This is the density of the second-highest value of $N$ draws.
You can rewrite this as
$$f_2^N(x) = N (1-F(x)) [(N-1)F^{N-2}(x)f(x)] = N (1-F(x)) f_1^{N-1}(x).$$
You can learn more about order statistics in Appendix C of Krishna's book and the references therein.
Why is this not exactly your equation with another $N$? Because your equation gives us something else. It is the density of the second-highest of $N-1$ iid draws conditional on the event that the highest of those draws is at most $x$.
The first part in the equation corrects for this event. Appendix A or any book on probability theory should give you an idea on conditional CDFs and densities. The second part is the analogue to $N(1-F(x))$ with another $N$ and also corrected for $Y_1<x$. The third part is also analogous to the final part of my equation above.
