Are the following two inner products on differential forms equal? There are two inner product on differential forms:

*

*$\langle \alpha,\beta\rangle$ induced from Riemannian metric $g$ by defining on 1-forms as dual of vector fields then extending to all differential forms i.e. $\langle e_{i_1}\wedge \dots\wedge e_{i_k}, e_{j_1}\wedge \dots\wedge e_{j_k}\rangle =\det[\langle e_{i_s}, e_{j_s}\rangle]$.

*On compact oriented Riemannian manifold $(M,g)$ $$(\alpha,\beta)=\int_M\alpha\wedge\star\beta.$$
Q1: Are these two inner products on differential forms equal?
Q2: If the answer to Q1 is "NO" then is it important to notice that two operators are adjoint of each other (or an operator is symmetric or self-adjoint)  w.r.t. which metric? e.g. $d$ and $\delta$ that are adjoint w.r.t. second inner product but I don't know it is w.r.t. other one.
 A: Disclaimer: the comment section is overgrowing so here is an answer that I hope will erase all your doubts.
First part:
As said in the comment section, $(1)$ is a inner product on the vector space $\Lambda^p_xM$ while $(2)$ is an inner product on the vector space $\Omega^p(M)$. The link between them is that $(2)$ is obtained by integrating $(1)$ over the whole manifold $M$.
An analog is this: consider the set $M=[0,1]$. Then for each $x\in [0,1], T_x[0,1] = \mathbb{R}$ and one can define an inner product on $T_x[0,1] = \mathbb{R}$ by $\langle a,b \rangle_x = a\times b$. This is $(1)$.
A vector field on $[0,1]$ is just a smooth function $f:[0,1] \to \mathbb{R}$, and $(2)$ is here
$$
\langle f,g\rangle = \int_0^1 f(x) g(x) \mathrm{d}x = \int_0^1 \langle f, g\rangle_x \mathrm{d}x.
$$
Second part:
If $V$ is a vector space and $\langle\cdot,\cdot\rangle$ is an inner product, one can create a Riemannian metric on $V$, thought as a manifold the following way. As a vector space, the tangent bundle of $V$ is trivial:
$$TV = V\times V$$
and one can define the Riemannian metric $g_v = \langle\cdot,\cdot\rangle$ for $v\in V$. It is a constant Riemannian metric because the canonical trivialization makes $g_v$ to be a function independant of $v \in V$.
Take $V = \Omega^p(M)$ and $\langle \alpha,\beta\rangle = \int_M \alpha \wedge \star \beta$. Now, forget that $V$ and $\langle\cdot,\cdot\rangle$ is defined thanks to a Riemannian manifold $(M,g)$ and just look at its structure: it is a vector space with an inner product. Hence, for this inner product $\|\alpha\|$ is a number.
If you really want to think of this construction as a Riemannian manifold, like in the first paragraph, then $\|\alpha\|$ will be a function:
$$
\|\alpha\| : \beta \in \Omega(M)^p \mapsto \|\alpha\|(\beta) = \|\alpha\|\in \mathbb{R} 
$$
which is constant and does not take points of $M$ as entries.
Comment: if you really do not understand what I said, here is just a question for you: for $x \in M$, how would you define $\left(\int_M \alpha\wedge \star \beta\right)(x)$?
This is the exact same thing as this question: how would you define $\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$?
