Write the torus as a closed square with opposite edges identified.
Take the square's centre as the point to be removed. Deform the
square minus its centre radially to its perimeter. When we identify the edges
this becomes a deformation retraction of the torus minus a point to a bouquet
of two circles.
Let $F \subset \mathbb R^2$ be a finite set. We want to construct a homeomorphism $f: \mathbb R^2 \rightarrow \mathbb R^2$ mapping $F$ to a fixed finite set of size $n$. We will adapt the convention that "applying a map $f$ to a point $p$" means that the letter $p$ now denotes what was previously $f(p)$. This gives the following procedure an algorithmic feel, which is easier to follow I hope.
First, note that we can ensure that one of the elements of $F$ is zero by just applying a translation. Once we are in this situation, let $p \in F$. Then applying a rotation (around zero) arranges that $p$ gets mapped to the $x$-axis. Applying a transformation of the form $(x,y) \mapsto (x, \lambda y)$ for big enough $\lambda$ ensures that all $p \in F \setminus x\text{-axis}$ have modulus strictly greater than all $q \in x\text{-axis}$.
Now comes the key step. Let $R_{\alpha}: \mathbb R^2 \rightarrow \mathbb R^2$ be a rotation (around zero) of angle $\alpha$ mapping a point $p \in F \setminus x\text{-axis}$ of minimal modulus to the $x$-axis. Then let $f: \mathbb R^2 \rightarrow \mathbb R^2$ be given as
$$f(s) = \begin{cases}
s & |s| \leq \max_{q \in x\text{-axis}}|q| \\
R_{(L(|s|))\alpha}(s) & \max_{q \in x\text{-axis}}|q|< |s|<\min_{q \in F \setminus x\text{-axis}}|q| \\
R_\alpha & |s| \geq \min_{q \in F \setminus x\text{-axis}} |q|,
\end{cases}$$
where $L: \mathbb R \rightarrow R$ is a linear function with $L(\max_{q \in x\text{-axis}} |q|)= 0$ and $L(\min_{q \in F\setminus x\text{-axis}}|q|) = 1$. This $f$ can be visualised as some kind of rotation twisting only outside of a ball around zero. Applying this function and again some $(x,y) \mapsto (x, \lambda y)$ produces the same situation but with strictly more points on the $x$-axis.
Proceeding by induction, we obtain that there exists an $f: \mathbb R^2 \rightarrow \mathbb R^2$ mapping $F$ to the $x$-axis. All that is left to show is that there exists a homeomorphism $h: \mathbb R^2 \rightarrow \mathbb R^2$ mapping $F \subset \mathbb R$ to some fixed set, say $\{0,1, \dots, |F|-1\}$. But this is easy, just consider some piecewise linear function $h$ that does this, and then apply $(x,y) \mapsto (h(x), y)$ to obtain our desired final function mapping $F$ to a specific set, namely $\{(0,0), (0,1), \dots, (0, |F|-1)\}$.
Best Answer
Instead of choosing lines going through the removed points, consider $n-1$ lines which separate the points into $n$ chambers. Each such punctured chamber deformation retracts onto its boundary circle. (Push along the dotted arrows in the following picture.)
Then you are left with the 1-skeleton of this complex, which is precisely a wedge of $n+1$ circles. (The boundary of the big square, under the quotient map to the torus, is a wedge of two circles. There are $n-1$ circles inside the big square, which are the images of the lines delimiting the chambers under the quotient map, for a total of $2+(n-1) = n+1$ circles.)