Let $M$ a set and $\sim$ an equivalence relation. Let $\pi: M\longrightarrow M/_\sim$ the projection. Do we have that $\pi(\partial M)\supset \partial (M/_\sim)$ ? (where $\partial A$ denote the boundary). I think yes but I have problem to prove it.
My question behind is : if we know the boundary of $M$ can we easily find the boundary of $M/_\sim$ ?
On
The answer about the conclusion is no For example, take $M=\mathbb{R}^2$ and the relation is $x ~ y \Leftrightarrow x-y \in \mathbb{Z}^2$. The quotient space is $\mathbb{R}^2 / \mathbb{Z}^2 \cong \mathbb{S}^2$. Obviously that $\partial M= \emptyset$ because $M$ is open, and $\partial (M/\sim)=\mathbb{S}^2$. There aren't anything to conclude about $\partial (M/\sim)$ from $\partial M$
On
If $f:N_1\to N_2$ is a homeomorphism, an $M/_\sim =M/f$, then $$\partial (M/_\sim)=\partial M\backslash (N_1\cup N_2).$$
In your case if you set $$N_1=\{(0,y)\mid y\in [0,1]\}=\{0\}\times [0,1]\quad \text{and}\quad N_2=\{(1,y)\mid y\in [0,1]\}=\{1\}\times [0,1].$$
Then $$\partial (M/_\sim)= ([0,1]\times \{0\})\cup( \{1\}\times [0,1])\cup ([0,1]\times \{1\})\cup (\{0\}\times [0,1])\backslash (N_1\cup N_2)$$ $$=([0,1]\times \{0\})\cup([0,1]\times \{1\})\cong \mathbb S^1.$$
I made a slight typo in one of my comments. Let me summarize the discussion here and try to answer your question. Let's compare $\pi(\partial M)$ and $\partial(M/\sim)$.
If $M=[0,1]$ and we identify the endpoints, the quotient is $S^1$ and so the image of the boundary is not always contained in boundary of the quotient.
If $M=[0,1]$ again and we identify $x \sim 1-x$, call the quotient $N$, which is again a copy of $[0,1]$. But $1/2 \in M$ is in the preimage of the boundary of $N$, and clearly not a boundary point of $M$. So the image of the boundary does not always contain the boundary of the quotient.
Now, you gave an example of the Mobius band. Let $M=[0,1] \times [0,1]$. Set $(0,y) \sim (1,1-y)$ and let $N$ be the quotient. Then the boundary of $M$ is $\partial M = \{ (0,s) \}\cup\{ (1,s) \}\cup\{ (t,0) \}\cup\{ (t,1) \}$ and the boundary of $N$ is $\partial N = \{ [t,0]\} \cup \{[t,1]\}$, and you notice that $\pi(\partial M)$ contains $\partial N$. Why is that so? Well all of the points you identified lay in the boundary of $M$. It's almost tautological: either a boundary point of $N$ was unchanged by the identification, in which case it was a boundary point of $M$, or it was identified in some way to "become" a boundary point: but then it was in the boundary of $M$ in the first place. This failed in the above example because we identified lots of non-boundary points.
We could also compare the interiors. Pick any point in the interior of $M$, and you can find a little open set that is mapped identically by a restriction of the quotient map. So the interior of $M$ is mapped into the interior of $N$ when all the identifications are on boundary points.
Let me know if you have questions (or if I made any errors).