Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $\Bbb R^{n+1}$. We want to use this
as a basis for one of $T^{n+1}=T^n\times S^1$ into $\Bbb R^{n+2}$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_{n+1}>0$. Then let the embedding of $T^n$
be given by functions
$p\mapsto (\phi(p),\psi(p))$ where $\phi(p)\in\Bbb R^n$ and $\psi(p)\in(0,\infty)$. Now the embedding of $T^{n+1}$ into $\Bbb R^{n+2}$ by
$$(p,e^{it})\mapsto(\phi(p),\psi(p)\cos t,\psi(p)\sin t).$$
As $\psi(p)>0$ the last two coordinates determine $\psi(p)$ and $e^{it}$.
Yes, it's just polar coordinates. Define $\phi:B\to \mathbb R$ as follows:
If $x>0, y\ge0,$ then $\phi(x,y)=\tan^{-1}(y/x).$
If $x\le0,y>0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{\pi}{2}.$
If $x<0,y\le0$ then $\phi(x,y)=\tan^{-1}(y/x)+\pi.$
If $x>0,y\le 0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{3\pi}{2}.$
Then, $\phi$ is smooth on $B$. Now, fix a point $p=(x,y):x>0,y\ge0.$ Then, $d\phi: T_pB\to T_{\phi(p)}\mathbb R$ is given by
$(d\phi)_p=(\partial_x)_pdx+(\partial_y)_pdy=\frac{-y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy=\omega.$
The same formula for $\omega$ is obtained in the other quadrants of $\mathbb R^2\cap B$.
If $g$ is a closed $0$-form, then $dg=0$. Now, in local coordinates, $dg=\partial_xgdx+\partial_ygdy.$ As $dg$ is $identically$ zero in $B$, we have $dg(\frac{\partial}{\partial x})_p=(\partial_xg)_p=0$ for all $p\in B.$ Similarly, $(\partial_yg)_p=0.$ Now you can either use the hint or just observe that since both partial derivatives of $g$ vanish on the connected set $B$, in fact $g$ must be constant there.
If $\omega=df$ on $A$ then in particular $df-d\phi=\omega-\omega=0$ on $B$ so $f-\phi=c,$ some constant, on $B$. Using the hint, $\lim_{y\to 0^-}(f(1,y)-\phi(1,y))=f(1,0)+2\pi=c$ and $\lim_{y\to 0^+}(f(1,y)-\phi(1,y))=f(1,0)=c$, which implies that $2\pi=0,$ a contradiction.
But it's easier just to integrate both sides around the unit circle. That is, if $\omega=df$ then $\int \omega=2\pi$ by direct calculation, whereas $\int df=0$ by the FTC. Or you can argue that if $\omega=df$ then, $df =\frac{-y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy$ with $\partial f_x=\frac{-y}{x^2 + y^2}$ and $\partial f_y=\frac{x}{x^2 + y^2}$. But then, you get a contradiction because the mixed partials are not equal.
Best Answer
Let $(U, \varphi)$ be a chart on $T^2$, i.e. $U$ is an open subset of $T^2$ and $\varphi : U \to \mathbb{R}^2$ is a homeomorphism. Let $V$ be the open subset of $U$ such that $\varphi|_V : V \to B(0, 1)$ is a homeomorphism, i.e. $V = \varphi^{-1}(B(0, 1))$. Then $\varphi(\overline{V}) = \overline{\varphi(V)} = \overline{B(0, 1)}$ and $\varphi(\partial V) = \partial\varphi(V) = \partial B(0, 1) = S^1$.
Note that the quotient $\overline{B(0, 1)}/S^1$ is homeomorphic to $S^2$; let $\psi : \overline{B(0, 1)}/S^1 \to S^2$ be a homeomorphism. The composite $\psi\circ\varphi|_{\overline{V}} : \overline{V} \to S^2$ maps $\partial V$ to a single point, call it $p$, and $(\psi\circ\varphi|_{\overline{V}})|_V = \psi\circ\varphi|_V$ is a homeomorphism from $V$ to $S^2\setminus\{p\}$.
Now define $f : T^2 \to S^2$ by
$$f(x) = \begin{cases} \psi(\varphi(x)) & x \in \overline{V}\\ p & x \not\in \overline{V}. \end{cases}$$
Then $f$ is a continuous map. Furthermore, it has degree one .
More generally, we can use the same technique to construct a degree one map from any closed, connected, orientable $n$-manifold to $S^n$.