Step 1

I can give you a partial answer, which is you are on the right track, but multiplying the 97.5% bounds will not yield a 95% confidence interval. Your interval will be too large.

I think what you are trying to do is estimate two unknown proportions ${p}_{1}$ and ${p}_{2}$. You compute two estimates ${\hat{p}}_{1}$ and ${\hat{p}}_{2}$ from the data you observe. You compute confidence intervals for each one ${I}_{1}=[{a}_{1},{b}_{1}]$ and ${I}_{2}=[{a}_{2},{b}_{2}]$, respectively. So, $P({p}_{1}\in {I}_{1})=P({p}_{2}\in {I}_{2})=0.975$. Since, they are independent, you can consider the confidence set $S={I}_{1}\times {I}_{2}$, which, geometrically, is a square in the plane. Now, $P(({p}_{1},{p}_{2})\in S)=0.95$ since we have independence. This is equivalent to $P({p}_{1}\in {I}_{1},{p}_{2}\in {I}_{2})=P({p}_{1}\in {I}_{1})P({p}_{2}\in {I}_{2})$. That is, the probability of this ordered pair being inside the square S is 95%. This is a way to get that 95% confidence set.

Step 2

However, if you are interested in a confidence interval for $p={p}_{1}{p}_{2}$ multiplying the bounds together to get an interval $I=[{a}_{1}{a}_{2},{b}_{1}{b}_{2}]$ will cause a problem. Using the same notation, you are asking for the probability $P(p\in I)$. Putting some concrete numbers to this, let ${I}_{1}={I}_{2}=[0.5,0.6]$, so that $I=[0.25,0.36]$. All pairs $({p}_{1},{p}_{2})\in [0.5,0.6{]}^{2}$ yield a $p\in [0.25,0.36]$, so $P(p\in I)\ge 0.95$. The problem is that there are also pairs $({p}_{1},{p}_{2})\notin [0.5,0.6{]}^{2}$ that have $p\in [0.25,0.36]$ such as $({p}_{1},{p}_{2})=(0.3,1.0)$ where $p=0.3$. Hence, $P(p\in I)>0.95$. Your confidence interval is too big.

In general, it will require more effort to produce a confidence interval for the product ${p}_{1}{p}_{2}$. Even if the your data is coming from nice distributions (e.g. normal, binomial) the product distribution will probably be much nastier. Off the top of my head, I do not have a good answer to how to produce such an interval.