One good place to start is to consider what happens when $P \subset Q$ or $Q \subset P$. The cases are basically the same, so let's assume that $P \subset Q$. Since $P \cap Q = (0)$, we must have that $P = (0)$, and so $R$ is an integral domain. Since any integral domain that is not a field has at least two distinct prime ideals, the problem then tells us that any integral domain that is not a field can be embedded in the product of two fields. That should lead you to believe that whatever construction we use is a very, very general way of obtaining fields and field-like things from integral domains. The obvious tool is to use the field of fractions!
So for an integral domain, the answer is now clear: Take the map $R \mapsto \operatorname{Frac}(R) \times \operatorname{Frac}(R)$, where I am using $\operatorname{Frac}(R)$ to denote the field of fractions of $R$, and the map can either put $R$ in the first copy, or the second, or the diagonal if you want.
What if $R$ is not an integral domain? In that case you have two ways to make it into one: quotient by $P$ or quotient by $Q$. Since $P \cap Q = 0$, the map $R \mapsto R / P \times R / Q$ has $(0)$ kernel, so $R$ is isomorphic to a subring of $R / P \times R / Q$. Now, $R / P$ and $R / Q$ might not be fields, but no matter, because they are integral domains! So we can just put $R / P \times R / Q$ into $\operatorname{Frac}(R / P) \times \operatorname{Frac}(R / Q)$, and composing the two maps together we get an injective ring homomorphism $R \to \operatorname{Frac}(R / P) \times \operatorname{Frac}(R / Q)$. The latter is the product of two fields, and $R$ is isomorphic to its image, which is a subring.
I suspect that this is beyond your background right now, but as inspiration for future studies, I'll point out that this is a completely natural thing to do in algebraic geometry: geometrically, a product of two fields is two closed points, and the condition that $R$ has two prime ideals whose intersection is $0$ means that $R$ is geometrically given by the union of two irreducible components. The map I described in my answer corresponds to taking each of the closed points of $\kappa(P) \times \kappa(Q)$ to the corresponding irreducible component of $R$, where $\kappa(P)$ is the residue field at $P$, and same for $Q$.
Again, don't worry if this doesn't mean anything to you right now, just know that if you learn enough algebraic geometry, facts like this will begin to feel "natural".
