How do we know that a surface minus finite number of points is homotopy equivalent to a bouquet of circles? In this post (Homotopy Equivalence of Punctured Tori), the author of the first answer states that a surface minus finite number of points is homotopy equivalent to a bouquet of circles. However, it doesn't seem very clear to me.

Is there any source I can look at for the proof or explanation?

 A: The case $g=0$ being straightforward, let me focus to the case $g \geq 1$.
Let $X=\Sigma_g -\{p_1, \ldots p_k\}$ be the topological space obtained by removing $k$ distinct points from a closed surface of genus $g \geq 1$. By Seifert-Van Kampen theorem, a presentation for the fundamental group of $X$ is $$\pi_1(X)=\langle a_1, \, b_1, \ldots a_g, \, b_g, \, c_1, \ldots, c_k \; | \; c_1c_2 \ldots c_k \Pi [a_i, \, b_i]=1 \rangle \simeq F_{2g+k-1},$$
namely, $\pi_1(X)$ is free on $2g+k-1$ generators.
On the other hand, by classical uniformization theory (look for instance at MO question 254687), the universal cover $\tilde{X}$ is homeomorphic to the hyperbolic plane $\mathbb{H}^2$, that is contractible. This means that $X$ is an aspherical $\mathrm{CW}$-complex, in particular its homotopy type is detected by its fundamental group alone.
Therefore $X$ is homotopically equivalent to a bouquet $\bigvee_{i=1}^{2g+k-1} S^1$ of $2g+k-1$ circles, since this is another aspherical $\mathrm{CW}$-complex with the same fundamental group as $X$.
