Fundamental group and Euler Characteristic of non orientable genus g surface with k points removed.

Question

I saw this problem when prepping for my prelim, what I this is to mimic the proof for fundamental group of the orientable case. Let $X$ be a non orientable surface of genus g with k points removed. I assumed all k points are in the first $\mathbb{R}\mathbb{P}^{2}$ and using Van-Kampen I defined $U:= \mathbb{R}\mathbb{P}^{2}-\{x_{1},...x_{k+1}\}$, $V := N_{g-1}-\{y_{1}\}$, where $N_{g}$ stands for non orientable genus g surface, and $U \cap V = S^{1}$. So in this case $U$ would be homotopically equivalent to $\vee^{k+1}S^{1} $ and $V$ to $\vee^{g-1}S^{1}$. Therefore $\pi(X) \cong \ast^{k+g-1} \mathbb{Z}$. But I am not entirely sure if what I did is correct. For the Euler characteristic, I saw this https://mathoverflow.net/questions/376021/how-do-we-know-that-a-surface-minus-finite-number-of-points-is-homotopy-equivale but I don't know if I can conclude that X is aspherical as they did in the orientable case and conclude that $\chi(X) = 2-k-g$. Thanks in advance for any help or observation.