Let’s work purely at the level of vector spaces. The manifold case follows by doing this pointwise at each tangent space.
Let $V$ be a finite-dimensional vector space over $\Bbb{R}$, and $g:V\times V\to\Bbb{R}$ a bilinear symmetric non-degenerate map. As you may know, this gives rise to a linear isomorphism $g^{\flat}:V\to V^*$, $g^{\flat}(v)=g(v,\cdot)$ (this is essentially the definition of non-degeneracy). The pair $(V,g)$ is a pseudo inner-product space. Now, by transporting structure via $g^{\flat}$, we get a pseudo inner-product space $(V^*,\tilde{g})$; explicitly, $\tilde{g}=(g^{\flat})_*g$, which even more explicitly means $\tilde{g}(\cdot,\cdot)=g(g^{\sharp}(\cdot),g^{\sharp}(\cdot))$.
The question now becomes whether it is possible to define a pseudo inner-product $T^r_s(g)$ on the various tensor spaces $T^r_s(V)= V^{\otimes r}\otimes (V^*)^{\otimes s}$ (hopefully you recognize the special cases $T^1_0(g)=g$, and $T^0_1(g)=\tilde{g}$; also it’s common to write simply $\langle\cdot,\cdot\rangle_g$, letting context distinguish the meaning, instead of $T^r_s(g)$ to mean the pseudo inner-product induced by $g$ on the tensor spaces). The answer is yes.
To define the pseudo inner-product on $T^r_s(V)$, we’ll invoke the universal property of tensor products. Set $W=V^{r}\times (V^{*})^{s}$ (Cartesian product, not tensor product), and consider the map $W\times W\to\Bbb{R}$,
\begin{align}
&\left((v_1,\dots, v_r,\alpha^1,\dots,\alpha^s),(w_1,\dots, w_r,\beta^1\dots,\beta^s)\right)\\
\mapsto &g(v_1,w_1)\cdots g(v_r,w_r)\cdot \tilde{g}(\alpha^1,\beta^1)\cdots \tilde{g}(\alpha^s,\beta^s).
\end{align}
This map, is clearly bilinear $W\times W\to\Bbb{R}$, and it is also $2(r+s)$-fold multilinear. Hence, by the universal property of tensor products, this map descends uniquely to a bilinear map $T^r_s(g):T^r_s(V)\times T^r_s(V)\to\Bbb{R}$ such that its action on pure tensors is as above:
\begin{align}
&(T^r_sg)\left(v_1\otimes\dots\otimes v_r\otimes\alpha^1\otimes\dots\otimes\alpha^s, w_1\otimes\dots\otimes w_r\otimes\beta^1\otimes\dots\otimes\beta^s\right)\\
=&g(v_1,w_1)\cdots g(v_r,w_r)\cdot \tilde{g}(\alpha^1,\beta^1)\cdots \tilde{g}(\alpha^s,\beta^s).
\end{align}
Checking non-degeneracy of $T^r_s(g)$ is possible, but it’s slightly painful.
So, in the special case of $r=0,s=2$, $T^0_2g$ is the unique bilinear map on $T^0_2(V)$ such that its effect on the pure $(0,2)$ tensors is $(T^0_2g)(\alpha^1\otimes\alpha^2,\beta^1\otimes\beta^2)=\tilde{g}(\alpha^1,\beta^1)\cdot \tilde{g}(\alpha^2,\beta^2)$.
If you prefer in index notation, then for each pair of indices you find in the matching location, you contract with $g$ or $\tilde{g}$: for any $(r,s)$ tensors $A,B$ over $V$,
\begin{align}
(T^r_sg)(A,B)=g_{a_1b_1}\cdots g_{a_rb_r}g^{\mu_1\nu_1}\cdots g^{\mu_s\nu_s}A^{a_1\dots a_r}_{\,\,\,\,\quad\mu_1\dots\mu_s} B^{b_1\dots b_r}_{\,\,\,\,\quad\nu_1\dots\nu_s},
\end{align}
or specializing to the $(0,2)$ case,
\begin{align}
(T^0_2g)(A,B)&=g^{\mu_1\nu_1}g^{\mu_2\nu_2}A_{\mu_1\mu_2}B_{\nu_1\nu_2}.
\end{align}
So, if you like the index raising/lowering game, then you can write this as $A^{\mu_1\mu_2}B_{\mu_1\mu_2}$.
Best Answer
The question asks you to show
Watch out, "$\mathrm{grad}f|_p$ is orthogonal to $\Sigma$, the level set through $p$" does not mean, as you seem to believe, the following:
There are two reasons why this does not even make sense:
What it really means is:
Below is a proof of this correct interpretation of the question.
You do not need to use any computation, only the definition of the gradient of a function and the characterisation of tangent vectors as derivatives of curves.
Let $p\in M$ be a regular point for $f\colon (M,g)\to \Bbb R$, and let $\Sigma = f^{-1}(f(p))$ be the level set through $p$. Since $p\in \Sigma$ is a regular point, $\Sigma$ is a smooth hypersurface in a neighbourhood of $p$. Let $u\in T_p\Sigma$ be a tangent vector, and $\gamma\colon (-\varepsilon,\varepsilon)\to \Sigma$ a smooth curve such that $\gamma(0) = p$, $\gamma'(0) = u$. Then $$ \DeclareMathOperator{\grad}{grad} \newcommand{\d}{\mathrm{d}} \langle \grad\! f|_p,u\rangle = \d_pf(u) = \left.\frac{\d \left(f\circ\gamma(t)\right)} {\d t}\right|_{t=0}. $$ Since $\gamma$ has range in $\Sigma$ which is by definition the level set through $p$, $f\circ \gamma\colon (-\varepsilon,\varepsilon) \to \Bbb R$ is constant, equal to $f(p) \in \Bbb R$. Consequently, one has $\langle \grad\! f|_p,u\rangle = 0$. This being true for any $u\in T_p\Sigma$, one concludes that $\grad f|_p \perp T_p\Sigma$.