For a function $f : X \to \mathbb R$ let $s(f) = \{ x \in X \mid f(x) \ne 0 \}$. Then the support of $f$, $\text{supp} f$, is defined as the closure of $s(f)$ in the space $X$. To answer your question it will be easier to work with $s(-)$ than with $\text{supp}$.
For your function $f : \mathbb R \to \mathbb R$ we have $s(f) = (-\pi,\pi)$. Thus $s(\phi_{2m+1}) = ((m-1)\pi,(m+1)\pi)$ and $s(\phi_{2m}) = ((-m-1)\pi,(-m+1)\pi)$ because $\phi_i$ is obtained from $f$ by a shift of the variable $x$.
This shows that $\text{supp}\phi_i$ always has the form $[k\pi,(k+2)\pi]$ for some $k = k(i)$.
To verify that $\Sigma(x) = \sum \phi_i(x) = 1$ for all $x$, note that only the indices $i$ make a contribution $> 0$ for which $x \in s(\phi_i)$. We have $s(\phi_1) = (-\pi,\pi)$, $s(\phi_2) = (-2\pi,0)$, $s(\phi_3) = (0,2\pi)$, $s(\phi_4) = (-3\pi,-\pi)$, $s(\phi_5) = (\pi,3\pi)$, etc.
Case 1: $x = r\pi$ with $r \ge 0$. We have $r \pi \in s(\phi_i)$ precisely for $i=2r+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have $r\pi \in s(\phi_i) = ((m-1)\pi,(m+1)\pi)$ iff $r = m$, i.e. $i = 2r+1$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r \pi - r \pi)}{2} = 1$.
Case 2: $x = r\pi$ with $r< 0$. Then $x \in s(\phi_i)$ precisely for $i=-2r$. This is true because for $i = 2m+1$ with $m \ge 0$ we have $s(\phi_i) \subset (-\pi,\infty)$ and for $i = 2m$ with $m \ge 1$ we have $r\pi \in s(\phi_i) = ((-m-1)\pi,(-m+1)\pi)$ iff $r = -m$, i.e. $i = -2r$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r\pi + (-r)\pi)}{2} = 1$.
For the next two case recall that $\cos(y\pm\pi) = -\cos(y)$.
Case 3: $x \in (r\pi, (r+1)\pi)$ with $r \ge 0$. Then $x \in s(\phi_i)$ precisely for $i=2r+1$ and $r = 2(r+1)+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have
$$(r\pi, (r+1)\pi) \cap s(\phi_i) = (r\pi, (r+1)\pi) \cap ((m-1)\pi,(m+1)\pi) = \begin{cases} \emptyset & m \ne r, r+1 \\ (r\pi, (r+1)\pi) & m = r, r+1\end{cases}$$
Thus with $y = x + \pi r$ we have $\Sigma(x)= \phi_r(x) + \phi_{r+1}(x) = f(y) + f(y +\pi) = \frac{2 + \cos(y) + \cos(y + \pi)}{2} = 1$.
Case 4: $x \in (r\pi, (r+1)\pi)$ with $r < 0$. This is similar as case 3.
This is an exercise in integration by parts. Since the Levi-Civita symbol satisfies $\epsilon_{ijk} = -\epsilon_{kji}$,
$$\sum_i\int v_i(\nabla\times u)_i dx= \sum_{i,j,k}\int v_i \epsilon_{ijk}\partial_j u_k dx = \sum_{i,j,k}\int \epsilon_{kji} \partial_j v_i u_k dx = \sum_k \int u_k (\nabla\times v)_k dx$$
Which means the curl operator is self-adjoint.
You can also think of curl as left matrix multiplication by a skew-symmetric “matrix” of partial derivatives, which picks up one minus sign when transferring onto $v$, and differentiation is skew-adjoint (by integration by parts) which picks up another minus sign, hence making curl self-adjoint.
Best Answer
Consider the three forms $\psi = x_1 dx_2\wedge dx_3 \wedge dx_4$. Write $\phi : \mathbb T^3 \to \mathbb R^4$, $\phi = (\phi_1, \cdots, \phi_4)$. Then
\begin{align} \int_{\mathbb T^3} \phi^* \psi &= \int_{\mathbb T^3} \phi_1 d\phi_2 \wedge d\phi_3 \wedge d\phi_4 \\ &= \int_{\mathbb T^3} \epsilon^{abc} \phi_1 \partial_b \phi_2 \partial _c \phi_3 \partial _c\phi_4 \ \mathrm d x\ \mathrm d y\ \mathrm d z. \end{align}
On the other hand,
$$\int_{\mathbb T^3} \phi^* \psi = \operatorname{deg} (\phi) \int_{\mathbb S^3} \psi, $$
where $\operatorname{deg}$ is the deg of the map $\phi$, which is an integer. Finally, by Stokes theorem,
$$\int_{\mathbb S^3} \psi = \int_B d\psi = \int_B dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4.$$
The last term is the volume of the unit ball in $\mathbb R^4$ and is $\pi^2/2$. Thus your term equals $\operatorname{deg}(\phi)$.
The generalization to the higher dimensional case should be easy.
Edit To clarify, in general for two compact orientable $n$-dimensional manifold $M, N$, the degree of a smooth map $\phi : M\to N$ defined as $$ \int_M \phi^* \alpha = \operatorname{deg}(\phi) \int_N \alpha, \ \ \ \forall \alpha $$ is always an integer. I am following section 4 in Bott and Tu here. The above equality depends only on the cohomology class $[\alpha]$ instead of $\alpha$ itself. Thus we can assume $\alpha$ is a bump form support in a small open set around any point $q\in N$. Given a smooth $\phi$, let $q\in N$ be a regular value for $\phi$ (which exists by Sard's theorem). Then $\phi^{-1}(q)$ is a compact smooth submanifold of dimension $0$: that is, a finite set of points. Also there are open neighborhood of $q\in N$ so that $\phi : \phi^{-1}(B) \to B$ is a covering. Thus
$$ \int_M \phi^* \alpha = \int_{\phi^{-1}(B)} \phi^* \alpha = \sum (\pm 1) \int_B\alpha $$
This $\sum (\pm 1)$ is the degree of $\phi$, you have $\pm 1$ since $\phi$ is a local diffeomorphism.