Your notation disconcerts me a little bit. I think there is a mistake somewhere.
First, your description of $I/\!\!\sim$ seems to me to be incorrect: what is on the right-hand side is what is called usually $I^2 = [0,1]^2 = [0,1]\times [0,1]$; the unit square of the plane $\mathbb{R}^2$. When you write $\text{"something"}/\!\!\sim$, you usually mean that your are quotienting out by some equivalence relation. I guess that, in our case, this relation is the one generated by
$$
(x,y) \sim (x',y') \qquad \Longleftrightarrow \qquad
\begin{cases}
x = x' & \text{if}\qquad y=0 \qquad\text{and}\qquad y'= 1 \\
y = y' & \text{if}\qquad x=0 \qquad\text{and}\qquad x' = 1\\
\end{cases}
$$
Then, I guess that you originally had the usual parametrization (the way topologists write it at least) of the torus in $\mathbb{R}^3$, $ f : [0,1]^2 \longrightarrow \mathbb{T}^2$, generated by rotating a circumference of radius $a$ along a circumference of radius $b$:
$$
f(x,y) = ((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) , b\sin (2\pi x ) )
$$
(If these $2\pi$ bother you, you can delete them, but then you have to change your unit square $[0,1]^2$ for $[0,2\pi]^2$.)
And then, your professor should have said something like: "Since this parametrization is compatible with the former equivalence relation $\sim $, it passes to the quotient inducing a well-defined map"
$$
\overline{f} : I^2/\!\!\sim \longrightarrow \mathbb{T}^2
$$
which is
$$
\overline{f}\overline{(x,y)} = f(x,y) =((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) , b\sin (2\pi x ) ) \ .
$$
Am I right?
Well, all this assumes that you previously knew that parametrization of the torus and, in this case, bijectivity is now obvious. But, just in case you haven't seen that $f$ before, look just at the two first components:
$$
((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) ) \ .
$$
Keeping $x$ fixed for a moment, this is a parametrization of a circumference of radius $a + b\cos (2\pi x) $: when $y$ goes from $0$ to $1$ in the unit interval $[0,1]$, we go round the circumference of centre $(0,0)$ and that radius. Right?
Ok, so let's assume that, for $(x,y), (x',y') \in [0,1]^2$ we could have
$$
(a + b\cos (2\pi x))\cos (2\pi y) = (a + b\cos (2\pi x'))\cos (2\pi y')
$$
and
$$
(a + b\cos (2\pi x)) \sin (2\pi y) = (a + b\cos (2\pi x')) \sin (2\pi y') \ .
$$
Since these are the coordinates of points in circumferences of radii $a + b\cos (2\pi x)$ and $a + b\cos (2\pi x')$, if they are the same point, these radii must be equal. Hence
$$
a + b\cos (2\pi x) = a + b\cos (2\pi x') \qquad \Longleftrightarrow \qquad \cos (2\pi x) = \cos (2\pi x') \ .
$$
Now, look at the third coordinate:
$$
b\sin (2\pi x) = b \sin (2\pi x' ) \qquad \Longleftrightarrow \qquad \sin (2\pi x) = \sin (2\pi x') \ .
$$
So, we would have both:
$$
\cos (2\pi x) = \cos (2\pi x') \qquad\text{and}\qquad \sin (2\pi x) = \sin (2\pi x')
$$
and, since $x,x' \in [0,1]$, if $x\neq x'$, this is only possible if one of them is $0$ and the other one is $1$.
Then we would get that
$$
\cos (2\pi y) = \cos (2\pi y') \qquad\text{and}\qquad \sin (2\pi y) = \sin (2\pi y')
$$
and conclude that also, if $y\neq y'$, then $y=0$ and $y'=1$ or vice versa.
So, our map $f$ is "almost" injective, except for these points on the perimetre of the unit square $[0,1]^2$ we have just found. But these points are the ones we are identifying with the equivalence relation $\sim$. Thus $\overline{f}$ is injective.
As for surjectivity: every point on $\mathbb{T}^2$ can be located saying in which meridian and parallel it lies, right? Well, this is exactly what you are doing when you say your point on the torus has coordinates $(x,y)$: the first one tells you the parallel and the second one the meridian. Varying $(x,y) \in [0,1]^2$ you go through all meridians and parallels.