Interesting, in the newest edition of Lee's "Riemannian Manifolds" this is Proposition 2.40, and perhaps it gives a nicer hint. Let's get a few preliminaries down.
On a Riemannian manifold $M$ we have a metric $g$. The inner product is given as
$$
\langle X,Y\rangle=g_{ij}X^iY^j
$$
for $TM$ with respect to the coordinate frame.
Similarly, by considering the canonical bundle isomorphism $\tilde{g}:TM\to T^*M$ given by the Riemannian metric, we may construct a bundle metric on $T^*M$ as follows. For any $\eta,\xi\in T^*M$, there exists unique $X,Y\in TM$ such that
$$
\eta=g(X,\cdot),\;\;\xi=g(Y,\cdot),
$$
and we then define
$$
\langle\eta,\xi\rangle=\langle X,Y\rangle.
$$
We also have that
$$
\langle\eta,\xi\rangle=g^{ij}\eta_i\xi_j
$$
with respect to the coordinate frame.
Given a local orthonormal frame $(E_j)$ for $TM$, it may be checked that the dual frame $(\phi^i)$ is actually a local orthonormal frame for $T^*M$.
Now consider the tensor bundle $T^k_\ell(M)$ and define the bundle metric as follows. For any (pure) $(\ell,k)$-tensors
\begin{align*}
X_1\otimes\cdots\otimes X_\ell\otimes\eta^1\otimes\cdots\otimes\eta^k \\
Y_1\otimes\cdots\otimes Y_\ell\otimes\xi^1\otimes\cdots\otimes\xi^k
\end{align*}
in $T^k_\ell(M)$ we define
$$
\langle X_1\otimes\cdots\otimes X_\ell\otimes\eta^1\otimes\cdots\otimes\eta^k,Y_1\otimes\cdots\otimes Y_\ell\otimes\xi^1\otimes\cdots\otimes\xi^k\rangle=\langle X_1,Y_1\rangle\cdots\langle \eta^k,\xi^k\rangle.
$$
We then extend this bilinearly and this defines a bundle metric on $T^k_\ell(M)$. Note, this is in the statement of the proposition, and is what I consider the "hint".
With this definition, we may check that it satisfies your definition in coordinates, hence it is equivalent. Finally, by the statement above about the local orthonormal frames, and the "linearity over the tensor product", you should be able to check that the tensor products of basis elements determines a local orthonormal frame for the tensor bundle.
Best Answer
Pull-back of tensor fields plays nicely with the tensor product, and pull-back also plays nicely with the exterior derivative (make sure you prove these facts first), and pull-back of real-valued functions is simply composition: $F^*g = g \circ F$. Also, yes, that is the general local expression for $A$ on $Y$.
Now, if we take a chart $(U,\phi = (x^1,\dots, x^m))$ on $X$, such that $F(U)\subset V$ (or atleast $F(U)\cap V \neq \emptyset$) then \begin{align} F^*A &= F^*(A_{j_1 \dots j_k}) \cdot F^*(dy^{j_1}) \otimes \cdots \otimes F^*(dy^{j_k}) \\ &= (A_{j_1\dots j_k} \circ F) \cdot d(y^{j_1}\circ F) \otimes \cdots \otimes d(y^{j_k}\circ F) \\ &= (A_{j_1\dots j_k} \circ F) \cdot \left[\dfrac{\partial (y^{j_1}\circ F)}{\partial x^{i_1}} dx^{i_1}\right] \otimes \cdots \otimes \left[\dfrac{\partial (y^{j_k}\circ F)}{\partial x^{i_k}} dx^{i_k}\right] \\ &= \left[(A_{j_1\dots j_k} \circ F) \cdot \dfrac{\partial (y^{j_1}\circ F)}{\partial x^{i_1}} \cdots \dfrac{\partial (y^{j_k}\circ F)}{\partial x^{i_k}} \right] dx^{i_1}\otimes \cdots \otimes dx^{i_k} \end{align} (throughout, the $j$'s range over $\{1,\dots, n\}$ and $i$'s range over $\{1,\dots, m\}$, and summation convention is used).
Using smoothness of $A$ and $F$, what can you now deduce about smoothness of $F^*A$?