Let $\alpha \in E^p$, then we can write $\alpha = H(\alpha) + (\alpha - H(\alpha))$. As $G(\alpha) \in (H^p)^\perp$ satisfies $\Delta G(\alpha) = \alpha - H(\alpha)$, we have
$$ ||G(\alpha)|| \leq C ||\Delta G(\alpha)|| = C ||\alpha - H(\alpha)|| \leq C||\alpha||,$$
thus $G$ is bounded. To show that it is self-adjoint, you need to restrict $G$ to $(H^p)^\perp$. Let $\alpha, \beta \in (H^p)^\perp$, $H(\alpha) = 0$ and by $\langle \alpha, \Delta \beta\rangle = \langle \Delta \alpha, \beta\rangle$ and changing $\alpha, \beta$ to $(\Delta |_{(H^p)^\perp})^{-1} \alpha$, $(\Delta |_{(H^p)^\perp})^{-1} \beta$ respectively, we have
$$\langle G(\alpha) , \beta\rangle = \langle \alpha, G(\beta)\rangle. $$
Thus $G$ is self adjoint. (Actually $G$ is also a compact operator)
As a preliminary, the tensor product is functorial, in the sense that it has a natural extension to linear maps. That is, if if we have linear maps between vector spaces $L_1:U_i\to V_1$ and $L_2:U_2\to V_2$, we may define $L_1\otimes L_2:U_1\otimes U_2\to V_1\otimes V_2$ as the unique linear extension of the expression $(L_1\otimes L_2)(u_1\otimes u_2)=L_1(u_1)\otimes L_2(u_2)$, and this product of maps several important properties, namely that the tensor product of isomorphisms is an isomorphism.
It's more convenient here to think of an inner product $g$ on a finite dimensional vector space $V$ as an isomorphism with the dual space $g:V\to V^*$ with inverse
$\tilde{g}:V^*\to V$. The functorality of tensor product ensures that there is a unique extension to controvariant tensor spaces that is compatible with tensor products, namely $g^{\otimes k}:V^{\otimes{k}}\to(V^*)^{\otimes{k}}$, which is also an isomorphism and thus an inner product. Taking appropriate tensor products of $g$ and $\tilde g$ allows one to construct similar isomorphisms on tensors of any signature. Going back to bilinear forms, this gives the proprty $g(v_1\otimes u_1,v_2\otimes u_2)=g(u_1,u_2)g(v_1,v_2)$.
It seems from the snippets you provide that Moroianu simply states this as a known result (usung $g$ for any such inner product, which I will also do), while Bleeker states the coordinate form, which can be derived without too much fuss. Letting $A$, $B$ be tensors of signature $(0,k)$,
$$\begin{align}
g(A,B)&=g\left(A_{i_1\dots i_k}e^{i_1}\otimes\dots\otimes e^{i_k},B_{j_1\dots j_k}e^{j_1}\otimes\dots\otimes e^{j_k}\right) \\
&=A_{i_1\dots i_k}B_{j_1\dots j_k}g\left(e^{i_1}\otimes\dots\otimes e^{i_k},e^{j_1}\otimes\dots\otimes e^{j_k}\right) \\
&=A_{i_1\dots i_k}B_{j_1\dots j_k}g(e^{i_1},e^{j_1})\dots g(e^{i_k},e^{j_k}) \\
&=g^{i_1 j_1}\dots g^{i_k j_k}A_{i_1\dots i_k}B_{j_1\dots j_k}
\end{align}$$
(the generalization to arbitrary signature is the same computation) The wedge product spaces are isomorphic to the spaces of alternating tensors, which are vector subspaces of the tensor spaces, so they inherit the inner product as a restriction of domain.
Another way of doing things is to construct the wedge product spaces independently. The wedge product is also functorial, so we can carry out exactly the same construction replacing $\otimes$ with $\wedge$, and obtain a similar family of inner products These will correspond up to a choice of normalization; in particular, the authors you cite chosen conventions such that the two differ by a factor of $k!$.
Best Answer
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 $\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$?