EDIT: There was some mistakes here, I meant to write $ [(\mathbb{S}^{1} \times \{ 1 \}) \times \{ 1 \}] $. Here is the right answer for $ C $: since $ f $ is a degree $ k$ map, we can homotope to the standard map $ f_k: e^{i \theta} \rightarrow e^{ i k \theta} $. This gives a homotopy $ S^1/f $ to $ S^1/f_k = S^1 $.
The second question is crucial. If you go around the circle in $ S^1 \subset \mathbb{T}^2 $ once, then the attaching map will take you around $ S^1 = \partial M $, $ k $ times over. When you go to homology, this goes to $ 2 k $, so overall $ [ \gamma ] \rightarrow 2k [ \gamma ] $. It will only go around the $ \mathbb{T}^2 $ once, so the map is just $ [ \gamma ] \rightarrow [ \gamma ] $.
I hope this will help!
Let $M_g$ be a genus $g$ surface, $g>1$.
If you had some map $f: T \to M_g$ inducing an injection on first homology, it would also induce an injection on fundamental groups. This is because the map on first homology is the abelianization of the map on fundamental groups, and $\pi_1(T)$ is already abelian.
So if this were the case, $\pi_1(M_g)$ would have a subgroup isomorphic to $\Bbb Z^2$. So by general covering space theory $M_g$ has a covering space with fundamental group $\Bbb Z^2$; as all covering spaces of a surface are surfaces, and (by classification of compact surfaces, and because fundamental groups of noncompact surfaces are free) the only surface with fundamental group $\Bbb Z^2$ is the torus, this means there is a covering map $T \to M_g$. Because $T$ is compact, this must be a finite-sheeted cover. But if $X \to Y$ is an $n$-sheeted cover of finite CW complexes (or manifolds, if you like), then $\chi(X) = n\chi(Y)$; and $\chi(T) = 0, \chi(M_g) = 2-2g$. This is a contradiction.
So no map $T \to M_g$ for $g>1$ induces an injection on first homology.
As a result of this, we can classify all maps $T \to M_g$ up to homotopy.
Because there is a unique map up to homotopy $K(\pi, 1) \to K(\pi',1)$ for every homomorphism $\pi \to \pi'$, and the universal cover of $M_g, g>0$ is homeomorphic to $\Bbb R^2$, $M_g$ is a $K(\pi,1)$. Because we know now that no map $\Bbb Z^2 \to \pi_1(M_g)$ is injective (this is the only part we couldn't have done beforehand!), $\pi_1(M_g)$ has no torsion (if it did, there's a surface quotient of $\Bbb R^2$ with finite fundamental group; but this surface must be noncompact so as above this is not possible), and the only torsion-free quotients (that aren't the 'trivial quotient') of $\Bbb Z^2$ are $\Bbb Z$ and the trivial group, the image of any homomorphism $\Bbb Z^2 \to \pi_1(M_g)$ is cyclic or trivial.
In the first case, we may factor the homomorphism as $\Bbb Z^2 \to \Bbb Z \to \pi_1(M_g)$, where the first map is the quotient by $\langle (a,b)\rangle$, where $\text{gcd}(a,b) = 1$. Such a map is represented by a map $T \to T \to S^1$ that sends $(a,b)$ to $(0,1)$ and then projection onto the first factor of $T = S^1 \times S^1$. The map $S^1 \to M_g$, then, is just the map representing some element of $\pi_1(M_g)$.
So the homotopy classes of maps $T \to M_g$ are the trivial map, and those that factor as above.
Best Answer
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.