Take a point $a\in \pi^{-1} b$. Take vectors $\tilde w_i$ and $ w_i'$ on $T_a E$ such that $d_a \pi(\tilde w_i)=d_a \pi( w_i') = v_i$. Notice $\tilde w_i - w_i' \in \ker d_a \pi= T_a \pi^{-1}b$.
Since the dimension of $\pi^{-1}b$ is $m$, the vectors $v_1,\ldots,v_m, \tilde w_1 - w_1'$ must be linearly dependent and, therefore,
$$\alpha(v_1,\ldots,v_m,\tilde w_1 - w_1',\tilde w_2,\ldots,\tilde w_{k-m})=0.$$
By the above identity we have
$$\alpha(v_1,\ldots,v_m,\tilde w_1,\tilde w_2,\ldots,\tilde w_{k-m}) = \alpha(v_1,\ldots,v_m,w_1',\tilde w_2,\ldots,\tilde w_{k-m}).$$
Analogously, we may exchange $\tilde w_2$ by $w_2'$, $\tilde w_3$ by $w_3'$, etc.
Therefore,
$$\alpha(v_1,\ldots,v_m,\tilde w_1,\tilde w_2,\ldots,\tilde w_{k-m}) = \alpha(v_1,\ldots,v_m,w_1', w_2',\ldots,w_{k-m}').$$
Here is a proof.
Consider $p\colon (0,2\pi) \times (-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb S^2$ the parametrization given by $p(\theta,\phi) = (\cos\theta\cos\phi,\sin\theta\cos\phi,\sin\phi)$. First, note that $p^*(i^*\omega)= (i\circ p)^* \omega$, and therefore,
\begin{align}
p^*(i^*\omega) &= p^*i^*(x dy\wedge dy + y dz\wedge dx + z dx \wedge dy)\\
&= (x(i\circ p)) (i\circ p)^* (dy\wedge dz) + (y(i\circ p)) (i\circ p)^*(dz \wedge dx) + (z(i\circ p)) (i\circ p)^* (dw\wedge dy)
\end{align}
Now, use the fact that the pullback and the wedge product commute; so that
\begin{align}
(i\circ p)^* (dy\wedge dz) &= ((i\circ p)^*dy)\wedge((i\circ p)^* dz)\\
(i\circ p)^* (dz\wedge dx) &= ((i\circ p)^*dz)\wedge((i\circ p)^* dx)\\
(i\circ p)^* (dx\wedge dz) &= ((i\circ p)^*dx)\wedge((i\circ p)^* dy)
\end{align}
Note that
\begin{align}
(x(i\circ p)) &= x\circ i \circ p = \cos\theta\cos\phi\\
(y(i\circ p)) &= y\circ i \circ p = \sin\theta\cos\phi\\
(z(i\circ p)) &= z\circ i \circ p = \sin\phi
\end{align}
and now, use the chain rule in order to show that
\begin{align}
(i\circ p)^*dx &= dx \circ d(i\circ p) = d (x\circ i \circ p) = d(\cos\theta\cos\phi)\\
(i\circ p)^*dy &= dy \circ d(i\circ p) = d (y\circ i \circ p) = d(\sin\theta\cos \phi)\\
(i\circ p)^*dz &= dz \circ d(i\circ p) = d (z\circ i \circ p) = d(\sin\phi)
\end{align}
Expand these equalities, e.g $d(\cos\theta \cos \phi) = -\sin\theta \cos \phi d\theta - \cos \theta \sin \phi d\phi$.
Gluing these equalities all together and using the fact that $\cos^2 + \sin^2 = 1$, you should find
$$
p^*(i^*\omega) = \cos\phi d\theta \wedge d\phi
$$
Since the complementary of $Im(p)$ in $\Bbb S^2$ has measure zero, and since $p$ is a diffeomorphism onto its image, it follows that
\begin{align}
\int_{\Bbb S^2} i^* \omega &= \int_{Im(p)} i^*\omega \\
&= \int_{(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})}p^*(i^* \omega)\\
&= \int_{(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})} \cos \phi d\theta\wedge d\phi\\
&:= \int_0^{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \phi d\theta d\phi
\end{align}
The last equality being true by definition of the integral of the top form $\cos\phi d\theta \wedge d\phi$ in the oriented manifold $(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})$.
Best Answer
For your first question about why integration along the fiber is important, note that it gives a cochain map of degree $-n$, $$\pi_*: A^k(M) \to A^{k-n}(B).$$ If $M$ was an oriented rank $n$ vector bundle over $B$ then it can be shown that this map induces isomorphism at the level of cohomology, $H^k(M) \to H^{k-n}(B)$. This is called the Thom isomorphism. You can find a proof in Bott & Tu's classic.
Now for your second part, notice that (1.15) is the defining property for $\pi_*\alpha$. That is if there were any other form $\omega$, which satisfied (1.15): $\int_B \omega\wedge\beta = \int_M \alpha\wedge\pi^*\beta$ for all $\beta$ (of appropriate degree) on the base, then we would have $\omega = \pi_*\alpha$. So, (1.15) uniquely characterizes $\pi_*\alpha$.
T show that $\pi_*(\alpha\wedge\pi^*\beta) = \pi_*\alpha\wedge\beta$, it suffices to show that $\pi_*\alpha \wedge \beta$ indeed satisfies the defining property (1.15) for $\alpha\wedge\pi^*\beta$. That is we need to show that, for any $\phi \in A^*(B)$.
$$ \int_B (\pi_*\alpha\wedge\beta)\wedge\phi= \int_M (\alpha\wedge\pi^*\beta)\wedge\pi^*\phi $$
but this is obvious because of the defining property of $\pi_*\alpha$ we have, $$ \int_B (\pi_*\alpha\wedge\beta)\wedge\phi = \int_B \pi_*\alpha\wedge(\beta\wedge\phi) = \int_M \alpha \wedge \pi^*(\beta\wedge\phi) = \int_M (\alpha \wedge \pi^*\beta)\wedge\pi^*\phi. $$
Lastly, regarding your comment to @Ted about references, I would suggest Bott & Tu's book. They define integration along the fiber more explicitly using local charts (similar to what Ted was suggesting you to do). Then both (1.15) and (1.16) follow easily from that definition. FWIW, these formulas together are called the Projection Formulas (see Proposition 6.15 in Bott & Tu).