You are mixing the geometer's convention with the analyst's convention.
The geometer's convention is, in $\mathbb{R}$, simply to take second derivative. Its spectrum is nonpositive (as witness by trigonometric functions), and in usual $\mathbb{R}^n$ gives (because of parallelizability no further information is encoded in the laplacian for $p$-forms) $\mathrm{div}\circ\mathrm{grad}$ (which also has nonpositive spectrum).
The second $\Delta=(\mathrm{d}+\mathrm{d}^*)^2$ is the analyst laplacian. It has nonnegative spectrum (as you can see from the square, and this is what analysts like because you can do things like log or arbitrary powers and whatnot unambiguously). In $\mathbb{R}^d$ it is $-\mathrm{div}\circ\mathrm{grad}$.
So $(fg)''=f''g+2f'g'+g''f$ translates to
- $\Delta(fg)=(\Delta f)g+2\nabla f\cdot\nabla g+f(\Delta g)$ for the geometer's laplacian; and
- $\Delta(fg)=(\Delta f)g-2\nabla f\cdot\nabla g+f(\Delta g)$ for the analyst's laplacian.
Here is an attempt of an answer.
Question 1 There is no need for an equality like that. What is true is that
$$
d\left(\alpha\wedge \gamma \right) = d\alpha \wedge \gamma + (-1)^{\deg\alpha}\alpha \wedge d\gamma
$$
And assuming your equality to be true will lead to an assumption on $d(\alpha\wedge\gamma)$
Here is a concrete counter-example:
\begin{align}
\alpha &= dx^1 & \gamma = x^2dx^3\wedge\cdots\wedge dx^n \\
d\alpha \wedge \gamma &= 0 & \alpha \wedge d\gamma = dx^1\wedge\cdots\wedge dx^n
\end{align}
Question 2 the answer is no. See above.
Question 3 above, the computations are local, so it does not depend on compactness or orientability: extend the counterexample by zero outside a chart.
Question 4 the answer is still no: in the counterexample above, $d\alpha\wedge \gamma = 0$, thus has zero integral, but $\alpha\wedge d\gamma$ is a volume form on an orientable manifold, it has non-zero integral.
Regarding @JanBohr's answer, (which leads to two self-refereing answers), I have to add that in case $M$ is oriented, then Stokes theorem states that
$$
\int_M d(\alpha\wedge \gamma) = \int_{\partial M} \alpha\wedge \beta
$$
and thus,
$$
\int_M d\alpha \wedge \gamma = (-1)^{\deg \alpha+1}\int_{M}\alpha\wedge d\gamma + \int_{\partial M}\alpha\wedge \gamma
$$
and thus there is (up to sign) an equality as soon as $M$ has no boundary or $\alpha\wedge \gamma$ is zero on $\partial M$.
Best Answer
First of all, I don't think there is any other reasonable interpretation using the slide besides that $$ \| \alpha \|^2 = (\alpha, \alpha).$$ This is used in no.34$\to$no.35 and no.39$\to$no.40. And
$$(\alpha, \alpha) = \int_M \alpha \wedge \star \alpha$$
are used when applying the formal adjoint $\delta$ in several places. So 1 is the interpretation used in the slide.
To show that $\|\alpha\| = 0\Rightarrow \alpha = 0$, we use 2.1, that is
$$ \alpha \wedge \star \alpha = g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$
so if $\|\alpha\|^2 = 0$, by 1 we have
$$0 = \int_M g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$
and since $g(\alpha, \alpha)$ is non-negative, $g(\alpha, \alpha) = 0$ everywhere and thus $\alpha(x) =0$ for all $x\in M$. So $\alpha = 0$.