Let $f$ be a ternary quadratic form, say with real coefficients. Then there is an associated symmetric matrix to $f$, say $A_f$. The orthogonal group of $f$ is then the group
$$\displaystyle O_f = \{H \in \operatorname{GL}_3(\mathbb{R}): H^T A_f H = A_f\}.$$
Given two non-proportional ternary quadratic forms $f,g$, which we may assume are not singular, how do we determine the group $O_f \cap O_g$? Is there a way to canonically write down generators of this group?
Think of a non-singular quadratic form as identifying the vector space with its dual. Given two, you can compose to get an automorphism of the vector space. The intersection of the orthogonal groups centralises that automorphism. As long as this automorphism is diagonalisable, you can simultaneously diagonalise the two forms. This occurs if, for example, one of them is positive definite or negative definite.
In this case, each eigenspace of the automorphism is preserved by the intersection, and the two forms are proportional there. Thus the intersection is a direct product of orthogonal groups acting on the eigenspaces of the automorphism.
If the automorphism is not diagonalisable, the centraliser is smaller and more complicated, but is still the intersection of the two orthogonal groups.
(These comments hold for arbitrary values of 3)
A good starting point might be A canonical form for a pair of real symmetric matrices that generate a nonsingular pencil by Frank Uhlig which gives a canonical form (over reals) of the pair $(A_f, A_g)$. Since you are happy with $3 \times 3$ matrices there are not that many cases to go through by hand.
The following complements the previous answers to arbitrarily many forms over any field $F$ of characteristic not $2$.
In general, given symmetric bilinear forms $f_1,\dots,f_t$ on $F^n$, the intersection $O_{f_1}\cap \dots \cap O_{f_t}$ will be of the form $O(A,\sigma):=\{a\in A:\sigma(a)a=1\}$ for some $F$-algebra $A$ and some $F$-linear involution $\sigma:A\to A$ that can be determined as follows.
Let $A$ be the subalgebra of $\mathrm{M}_n(F)\times \mathrm{M}_n(F)^{\mathrm{op}}$ consisting of pairs $(U,V^\mathrm{op})$ such that $$f_i(Ux,y)=f_i(x,Vy)\qquad\forall i,\,\forall x,y\in F^n$$ (in other words, $V$ is the dual of $U$ with respect to all the $f_i$). The involution $\sigma:A\to A$ is defined by $\sigma(U,V^{\mathrm{op}})=(V,U^{\mathrm{op}})$. Now observe that if $U\in O_{f_1}\cap\dots\cap O_{f_t}$, then $U^+:=(U,(U^{-1})^{\mathrm{op}})\in A$ and satisfies $\sigma(U^+)U^+=1_A$. This defines a group homomorphism $$O_{f_1}\cap \dots\cap O_{f_t}\to U(A,\sigma)$$ and it is a nice exercise to check that this is an isomorphism.
Back to the case of two forms $f,g$: Given matrices $U,V\in \mathrm{M}_n(F)$, the conditions $f(Ux,y)=f(x,Vy)$ and $g(Ux,y)=g(x,Vy)$ unfold to $U^t A_f=A_f V$ and $U^t A_g=A_g V$. Since $f$ is non-degenerate, $U$ determines $V$, i.e., $V=A_f^{-1}U^t A_f$, and thus $(U,V^\mathrm{op})\in A$ if and only if $V=A_f^{-1}U^t A_f$ and $U$ and $J:=A_f^{-1}A_g$ commute. Thus, there is an isomorphism $$ A\cong \mathrm{Cent}_{\mathrm{M}_n(F)}(J)=:C$$ given by $(U,V^{\mathrm{op}})\mapsto U$, and under this isomorphism, $\sigma$ is corresponds to the involution $\tau:C\to C$ given by $\tau(X)=A_f^{-1}X^t A_f=A_g^{-1}X^t A_g$. Thus, $$ O_f\cap O_g \cong U(C,\tau)=U(\mathrm{Cent}_{\mathrm{M}_n(F)}(J), X\mapsto A_f^{-1} X^tA_f). $$ It is further worth noting that $J=A_f^{-1}A_g\in C$ and is stable under $\tau$. Thus, if the minimal polynomial of $J$ has degree $n$ (which happens in the generic case), then $C=F[J]$ and $\tau=\mathrm{id}_C$, so $$ O_f\cap O_g=U(C,\mathrm{id}_C)=\{c\in C:\,c^2=1\}. $$ If it is moreover the case that the characteristic polynomial of $J$ has $n$ distinct roots, then $C\cong F\times\dots\times F$ and $$ O_f\cap O_g\cong \{\pm 1\}^n.$$ This is the "generic" behavior over an algebraically closed field.
For $3\times 3$ matrices over $\mathbb{R}$, it should be possible to list down all the possibilities for $C$ and $\tau$ and get a full classification.
Returning to the general case, the structure of the group $U(A,\sigma)$ is generally well understood. That $O_{f_1}\cap\dots \cap O_{f_t}$ has this form was probably known at least since the 1960s. For example, look at sections 2 and 3 of Bayer-Fluckiger and Lenstra's paper "Forms in Odd Degree Extensions and Self-Dual Normal Bases" and the references cited in this source.
