I believe you want to know why the VIX is a weighted portfolio of calls and puts with weights proportional to $\frac{1}{K^2}$ (NB: obviously the T is there to adjust for time to maturity, hence is not very enlightening).
Let's start with the basics. As shown by Breeden and Litzenberger (1978) the second derivative of an option price with respect to the strike is proportional to the risk-neutral probability, i.e. $\frac{\partial^2C}{\partial K^2}=\frac{\partial^2P}{\partial K^2} \propto q(K)$ where $q(K)$ is the risk-neutral pdf at $K$. The intuition is that an infinitely tight butterfly spread around $K$ gives us the risk-neutral likelihood of $K$.
Now that we have $q(K)$ we can construct in principle any desired payoff $f(S_T)$, given that $$price(f(S_T))=e^{-rT}E^Q[f(S_T)]=e^{-rT}\int_0^\infty f(K)q(K)dK=\int_0^\infty f(K) \frac{\partial^2C}{\partial K^2}dK$$ If you integrate by parts the previous integral and assume that the function is well behaved at the boundaries as shown here you obtain that:
$$price(f(S_T))=e^{-rT}f(F)+\int_0^F\frac{\partial^2 f}{\partial K^2}\left(K\right)P(K)dK+\int_F^\infty\frac{\partial^2 f}{\partial K^2}\left(K\right)C(K)dK$$ where $F$ is the current Forward price, $C$ and $P$ are call and put prices.
As shown by Neuberger, if prices follow a geometric Brownian motion, i.e. $dS=S\mu dt+S\sigma dW$, then $$log E^Q[S_T]-E^Q[log S_T]\propto\sigma^2$$ (NB: just write down the cumulant-generating function of a Normal distribution). This shows that under lognormal stock prices if we set $f(S_T)$ we can recover the implied volatility (NB: the implied risk-neutral variance, but the physical and risk-neutral variances are equal because of Girsanov's theorem in a continuous time diffusion.) Hence, you see immediately that to recover the price of the log contract you have to compute $\frac{\partial^2}{\partial K^2}log(K)=-\frac{1}{K^2}$. This is the reason why in the VIX$^2$ computation the weights are inversely proportional to $K^2$.
By the way, as a side note, if you are interested in risk-neutral variances outside of the log-normal paradigm it is enough to notice that $Var^Q(S_T)=E^Q[S_T^2]-E^Q[S_T]^2$, hence if you set $f(S_T)=S_T^2$ then you will notice that $\frac{\partial^2}{\partial K^2}K^2=1$ and therefore:
$$Var^Q(S_T)=...\int_0^F P(K)dK+\int_F^\infty C(K)dK$$