Does revealing shares of the result reveals shares of the inputs? Suppose Alive and Bob have shares of two polynomials $P,Q$, say, Alice knows $P_A,Q_A$, Bob knows $P_B,Q_B$ s.t. $P=P_A+P_B$ and $Q=Q_A+Q_B$. They perform some operation $f$ over the shared values, to obtain shares of $f(P,Q)$.

If they reveal the shares of $f(P,Q)$, will this give hints on their individual shares of $P,Q$?

I am supposing that one cannot deduce $P$ or $Q$ with the knowledge of $f(P,Q)$. I just want Alice and Bob to know $f(P,Q)$ without leaking shares.
It depends on how you make shares out of a polynomial. Consider for example Shamir secret sharing, using which a party can share a secret $s \in F$ (where $F$ is a finite field) with $n$ parties by doing the following:


Construct a polynomial $f(x) = s + a_1x+a_2x^2+\dots + a_nx^t$, for some $t$, where $a_i \in_R F$ for all $1 \leq i \leq n$.
Send the value of $f(i)$ to party $P_i$ for all $1 \leq i \leq n$.


The above mentioned secret sharing scheme is linear and it further allows us to perform multiplication of shared values as well. When using Shamir secret sharing for evaluating an arithmetic circuit securely, we release the shares of the final output in public, from which the final output can be constructed. The proof for the fact as to why the shares donot leak anything more than the described output can be found in the Chapter 3 of the textbook titled "Secure Multiparty Computation and Secret Sharing".