This is indeed a rather cumbersome calculation. Here it goes:
$$\nabla R(e_i,e_j,e_k,e_l,e_h)(p)+\nabla R(e_i,e_j,e_h,e_k,e_l)(p)+\nabla R(e_i,e_j,e_l,e_h,e_k)(p)=$$
$$\langle \nabla_{e_h} \nabla_{e_l} \nabla_{e_k}e_i - \nabla_{e_h} \nabla_{e_k} \nabla_{e_l}e_i +\nabla_{e_h} \nabla_{[e_k,e_l]}e_i$$
$$+\nabla_{e_l} \nabla_{e_k} \nabla_{e_h}e_i - \nabla_{e_l} \nabla_{e_h} \nabla_{e_k}e_i +\nabla_{e_l} \nabla_{[e_h,e_k]}e_i$$
$$+\nabla_{e_k} \nabla_{e_h} \nabla_{e_l}e_i - \nabla_{e_k} \nabla_{e_l} \nabla_{e_h}e_i +\nabla_{e_k} \nabla_{[e_l,e_h]}e_i, e_j \rangle (p) =$$
$$ \langle R(e_l,e_h)\nabla_{e_k}e_i-\nabla_{[e_l,e_h]}\nabla_{e_k}e_i+R(e_k,e_l) \nabla_{e_h}e_i -\nabla_{[e_k,e_l]}\nabla_{e_h}e_i+R(e_k,e_h)\nabla_{e_l}e_i-\nabla_{[e_k,e_h]}\nabla_{e_l}e_i+\nabla_{e_h} \nabla_{[e_k,e_l]}e_i+\nabla_{e_l} \nabla_{[e_h,e_k]}e_i+\nabla_{e_k} \nabla_{[e_l,e_h]}e_i,e_j\rangle (p)$$
$$= \langle R([e_l,e_h],e_k)e_i-\nabla_{[[e_l.e_h],e_k]}e_i+R([e_k,e_l],e_h)e_i-\nabla_{[[e_k.e_l],e_h]}e_i+R([e_k,e_h],e_l)e_i-\nabla_{[[e_k.e_h],e_l]}e_i,e_j\rangle(p) =$$
$$\langle R(\nabla_{e_l}e_h-\nabla_{e_h}e_l,e_k)e_i + R(\nabla_{e_k}e_l-\nabla_{e_l}e_k,e_h)e_i+R(\nabla_{e_k}e_h-\nabla_{e_h}e_k,e_l)e_i-
\nabla_{[[e_l.e_h],e_k]+[[e_k.e_l],e_h]+[[e_k.e_h],e_l]}e_i,e_j\rangle(p) = 0$$
In this last line the Jacoby Identity and the symmetry of the connection is used.
Observe that every time that elements of the form $\nabla_{e_l}e_h$ are dropped, what is being used is that of course $\nabla_{e_l}e_h(p)=0$, and the tensorial property of the Riemannian tensor (observe also the necessity that everything must be evaluated on p). Since p is arbitrary, and because of the linearity of everything involved, the identity is proven.
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 $\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$?
Best Answer
If $T$ and $S$ are type $(r,s)$ tensor fields on a pseudo-Riemannian manifold $(M,g)$, then $$\langle T,S\rangle = \sum g_{i_1k_1}\cdots g_{i_rk_r} g^{j_1\ell_1}\cdots g^{j_s\ell_s}T^{i_1\cdots i_r}_{\phantom{i_1\cdots i_r}j_1\cdots j_s}S^{k_1\cdots k_r}_{\phantom{k_1\cdots k_r}\ell_1\cdots \ell_s}.$$This is a definition (easily seen to be coordinate-independent). If the metric is positive and we have fixed a local orthonormal frame $(e_i)_{i=1}^n$ (and hence a metrically equivalent coframe $(\theta^i)_{i=1}^n$), then:
If you have an orthonormal frame, then $$\langle \nabla X, \nabla X\rangle = \sum \delta_{si}\delta^{tj} (\nabla X)^s_t(\nabla X)^i_j = \sum [(\nabla X)^i_j]^2,$$but what is $(\nabla X)^i_j$? By orthonormal expansion, we have that $\nabla_{e_j}X = \sum \langle \nabla_{e_j}X, e_i\rangle e_i$, so $(\nabla X)^i_j = \langle \nabla_{e_j}X,e_i\rangle$, and we can plug this into the above formula to get $$\begin{align} \langle \nabla X, \nabla X\rangle &= \sum [(\nabla X)^i_j]^2 = \sum \langle \nabla_{e_i}X,e_j\rangle \langle \nabla_{e_j}X,e_i\rangle \\ &= \sum \left\langle \nabla_{e_j}X, \sum\langle \nabla_{e_j}X,e_i\rangle e_i\right\rangle = \sum \langle \nabla_{e_j}X,\nabla_{e_j}X\rangle\end{align}$$
A $2$-form $\omega$ is just a very particular type $(0,2)$ tensor field, so $$\langle \omega,\omega\rangle = \sum g^{ik}g^{j\ell}\omega_{ij}\omega_{k\ell}$$and that's it. The covariant derivative $\nabla\omega$ is a type $(0,3)$ tensor field, so all of the above applies, and we have $$\langle \nabla \omega, \nabla \omega\rangle = \sum g^{ir}g^{js}g^{kt}\omega_{ij;k}\omega_{rs;t}.$$Particularizing this, we have $$\langle \nabla \omega,\nabla\omega\rangle = \sum \delta^{ir} \delta^{js}\delta^{kt} \omega_{ij;k}\omega_{rs;t} = \sum [\omega_{ij;k}]^2,$$but what is $\omega_{ij;k}$? By orthonormal expansion again, we have that $$\nabla_{e_k}\omega = \sum \langle \nabla_{e_k}\omega, \theta^i \otimes \theta^j\rangle \theta^i\otimes \theta^j,$$so $\omega_{ij;k} = \langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle$, leading to $$\begin{align} \langle \nabla \omega, \nabla \omega\rangle &= \sum [\omega_{ij;k}]^2 = \sum \langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle \langle \nabla_{e_k}\omega, \theta^i \otimes \theta^j\rangle \\ &= \sum \left\langle \nabla_{e_k}\omega, \sum\langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle \theta^i\otimes \theta^j\right\rangle =\sum \langle \nabla_{e_k}\omega, \nabla_{e_k}\omega\rangle.\end{align}$$
Let me know if you want me to be more careful with the bounds of summation, I was just using Einstein's convention in my mind (which apparently you're familiar with). If you want to consider the inner product induced in the exterior algebra as opposed to the tensor algebra, all of this should work the same (but you pay the price either with a $(\deg \omega)!$ somewhere or restricting the sum to increasing indices --- I'll let you pick your poison there).