I would think about the map in the opposite way:
When you said that the coordinates of the torus are $$((R+r\cos \theta)\cos \phi,(R+r\cos \theta)\sin \phi,r\sin \phi)$$ you were providing a global parametrization of the torus.
This is a map $$f(\theta, \phi) = (x(\theta, \phi), y(\theta, \phi), z(\theta, \phi))$$
With $x(\theta, \phi) = (R+r\cos \theta)\cos \phi$, etc. etc...
Each of the coordinate maps $x,y,z$ is continuous because they simply inherit continuity from the elementary trigonometric functions (finite products and sums of continuous functions are continuous), so the parametrization $f(\theta, \phi)$ is continuous. $f$ is also clearly bijective, as you noted.
Now all you need to show is that $f^{-1}$ is continuous. Equivalently, you said, you could show that $f$ is an open map. Or, better yet, since $f$ is a bijection, it's being an open map is equivalent to being a closed map.
Let's abstract for a moment. If you have any continuous $f: X \rightarrow Y$ with $X$ compact and $Y$ a $KC$-space (compact subsets of $Y$ are closed), then you have that $f$ is a closed map. This is very general and very important! Closed subsets of compact spaces are compact, and compactness is invariant under continuous maps. If compactness implies closedness in $Y$, then $f$ maps closed subsets of $X$ to closed subsets of $Y$ just because it's continuous. So, a function $f$ on a compact space is a homeomorphism iff it is bijective and continuous (so long as $Y$ has a bit of structure).
Now in your case $Y$ is $\mathbb{R}^3$, in which compact is equivalent to closed and bounded, so certainly $Y$ is a $KC$-space. Meanwhile recall that $S^1$ is compact and therefore $S^1 \times S^1$ is compact space since the product of two compact spaces is compact. So, having already seen that $f$ is a continuous bijection, $f: X \rightarrow Y$ is a homeomorphism.
The quotient space has a universal property, namely if $f:X\rightarrow Z$ is a continuous function with the property that $(p(a)=p(b))\Rightarrow (f(a)=f(b))$, then $f$ factors through the quotient space. In other words, there exists a unique continuous function $\widetilde{f}:Y\rightarrow Z$ so that $f=\widetilde{f}\circ p$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \ \searrow^{f} & \\
Y & \underset{\bar f}{\longrightarrow} & Z
\end{matrix}
$$
If you don't take the quotient map to be surjective, this property fails (the map fails to be unique).
From the definition you state, it seems that one could define a quotient even if map $p$ is not surjective, but this breaks a desired property (that might not have been mentioned in your studies yet).
If you consider a "quotient map" $q$ that has all the properties above, but isn't surjective, then you can write $q$ as a composition of a (true) quotient and an inclusion map. In other words, the composition $X\stackrel{p}{\rightarrow} Y\stackrel{i}{\hookrightarrow}Z$ equals $q$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \ \searrow^q & \\
Y & \underset{i}{\hookrightarrow} & Z
\end{matrix}
$$
Best Answer
This is an application of a standard theorem of topology which you can find, for example, in Munkres "Topology", which goes like this.
If $f : X \to Y$ is a continuous, surjective function from a compact space $X$ to a Hausdorff space $Y$, then $f$ is a quotient map (the proof is elementary, and uses the fact that a compact subset of a Hausdorff space is closed, and a closed subset of a compact Hausdorff space is compact).
Furthermore, let $Q$ be the quotient space of $X$ obtained by decomposing $X$ into the point pre-images $f^{-1}(y)$, $y \in Y$. Also let $F : Q \to Y$ be the induced function $F(f^{-1}(y))=y$. Then $F$ is a homeomorphism (again the proof is elementary, being just an application of the definition of a quotient map and the quotient topology).
In your example, $X = [0,1] \times [0,1]$ which is clearly compact, $Y$ is a subspace of $\mathbb{R}^3$ which is clearly Hausdorff, and $f$ is clearly continuous and surjective.