We have
$$w = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} \ ,$$
thus
$$dw = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} dx_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} = \frac{1}{r} \sum_{i=1}^{n+1} dx_1\cdots dx_{n+1} = \frac{n+1}{r} dx_1 \cdots dx_{n+1}\ .$$
Note that the second equality holds because in commuting those $dx_j$ we have the rule
$$dx_i dx_j = - dx_j dx_i\ .$$
Now by Stokes theorem, as $\mathbb S^n = \partial B^{n+1}$,
$$\int_{\mathbb S^n} w = \int_{\partial B^{n+1}} w = \int_{\mathbb B^{n+1}} dw = (n+1) V(B^{n+1})\neq 0$$
Thus $w$ when restricted to $\mathbb S^n$ is not exact (that it is not exact in $\mathbb R^{n+1}$ is obvious as $dw \neq 0$). The reason is again by Stokes theorem: if $w = d\alpha$, then
$$\int_{\mathbb S^n} w = \int_{\mathbb S^n} d\alpha = 0\ .$$
But we have seen that this is not zero.
I'll illustrate the method that can be used with a slightly different function from the one given in the question. For the purpose of this demonstration, I'm going to consider the function
$ f(\sigma, \phi) = C_p \sin(\sigma) \cos(\phi) $
Now, on the unit sphere, with the parameterization given and illustrated in the diagrams, we have
$x = \cos( \sigma)$
$y = \sin (\sigma )\sin (\phi)$
$z = - \sin( \sigma) \cos( \phi)$
Define the unit vector $u = [x, y, z]^T $, then
$C_p = 2 (v^T u)^2$ where
$v = [\lambda, \nu, \omega]^T$
Hence, my function is
$ f = 2 (v^T u)^2 (- z ) $
The surface integral is
$I =\displaystyle \iint_S f dS$
where $dS = \sin(\sigma) d\sigma d\phi$,
Since the region for the surface integral is the hemi-sphere that is pointed to by the vector $v$, then this suggests that we define a rotation of coordinates $r = R r'$ , where $R$ is a $3\times3$ rotation matrix whose columns point in the direction of the new axes. We will take the third column (the $z'$ axis) to be the unit vector along the vector $v$, while the first and second columns are arbitrary, as long as they together with the third column form a rotation matrix.
Written explicitly, the matrix $R = [u_1, u_2, u_3] $ where $u_1, u_2, u_3$ are the columns of matrix $R$, and we have $u_3 = \dfrac{v}{\|v\|} $
Since the $r = R r'$, then $r = x' u_1 + y' u_2 + z' u_3$. Where $r$ is the position vector in the original frame and $r'$ is the position vector in the new frame.
Now,
$v^T u = v^T R u' = (R^T v)^T u' = [0, 0, \|v\|] u' = \|v\| u'_z $
In terms of the new frame, points on the unit sphere are defined in terms of two angle $\theta, \psi$ as follows
$u' = \begin{bmatrix} \sin(\theta) \cos(\psi) \\ \sin(\theta) \sin(\psi) \\ \cos(\theta) \end{bmatrix} $
Therefore, $v^T u = \|v\| \cos(\theta) $
Recalling that the function we want to integrate is
$ f( \sigma, \phi ) = - 2 (v^T u)^2 z$
And from the expansion of $r$ in terms of $r'$, we get
$z = x' u_{13} + y' u_{23} + z' u_{33} = c_1 \sin( \theta) cos( \psi) + c_2 \sin (\theta) \sin(\psi) + c_3 \cos(\theta) $
Where $c_1 = u_{13} , c_2 = u_{23} , c_3 = u_{33} $
Hence, our function to be integrated over the hemisphere surface is
$ f(\theta, \psi) = -2 \|v\|^2 \cos^2(\theta) (c_1 \sin( \theta) \cos( \psi) + c_2 \sin (\theta) \sin(\psi) + c_3 \cos(\theta)) $
And the integral is
$ I = \displaystyle \large -\int_0^{2\pi} \int_0^\frac{\pi}{2} f(\theta, \psi) \sin(\theta) \text{d}\theta \text{d} \phi $
Integrating with respect to $\psi$ first, the terms involving $\cos(\psi)$ and $ \sin(\psi) $ cancel out, while the remaining term is multiplied by $2 \pi$
Hence, the integral is now
$ I = \displaystyle \large -(2\pi) \int_0^\frac{\pi}{2} 2 \|v\|^2 c_3 \cos^3(\theta) \sin(\theta) \text{d}\theta $
And this evaluates to
$ I = \large -\pi c_3 \| v \|^2$
And since $ v = [ \lambda, \nu, \omega ] $ , the above is simply
$ I = - \large \pi \omega ( \lambda^2 + \nu^2 + \omega^2 )^{\frac{1}{2}} $
Best Answer
A far easier approach is to apply Stokes's Theorem (in the other direction), noting $S^n = \partial(D^{n+1})$, and you get $$\int_{S^n}\omega = \int_{D^{n+1}}d\omega = (n+1)\text{vol}(D^{n+1}).$$