The problem: Show that $$F^\ast(AB)=(F^\ast A)(F^\ast B)$$ where $F$ is a smooth map from a smooth manifold $M$ to another smooth manifold $N$, A and B are symmetric tensor fields on $N$, and $F^\ast$ denotes the pullback where $$(F^\ast A)(v_1,...,v_k) = A(dF(v_1),...,dF(v_k))$$ gives its action on an element of the $k$th-power of a vector space (which in this case means the tangent space to $M$ at $p$, $T_pM$). (Exercise 12.27 in Lee's Introduction to Smooth Manifolds.)
Attempt at a solution: We have some identities like that $AB = \frac{1}{2}(A \otimes B + B \otimes A)$, and for writing $F^\ast A$ as a sum of basis tensors with the constants factored out (here, vector spaces are always Euclidean modules). This really hasn't gotten me very far as I'm not sure for instance how to take the tensor product of two sections of a tensor bundle.
I have that the RHS is equal to the product of a bunch of coefficients depending on $A$ and $B$ with the square of the sum $F^\ast (e)$, where $e$ ranges over all elements of a basis of the space of covariant tensors over $N$ in a neighborhood of $F(p)$. However this last term is also something I'm having trouble interpreting more concretely, or showing equal to the LHS.