A lot of the work on smooth manifolds is to let us use Euclidean analysis to merely locally Euclidean things that come up. Things that aren't Hausdorff are terrible and scary, real intuition busters (at least in my case), so I don't mind at all that we require that. And in fact, with just these 3 requirements (and a smoothness requirement), smooth manifolds can actually be viewed more or less in real space (the Whitney Embedding Theorem). And our real space intuition is pretty good, you know? And it suffices, has really nice properties, provides some pretty rich material, etc.
But there are some people who care a lot about non-Hausdorff manifolds. In fact, secretly, we see them and don't know it. The etale space of the sheaf of continuous real functions over a regular manifold is a manifold, and it's sometimes non-Hausdorff.
Similarly, there are people who care about non-second-countable manifolds. But these are also a bit unfortunate. One of the great things about second countability is that it guarantees that ordinary manifolds are paracompact. A paracompact smooth manifold admits partitions of unity subordinate to a refinement of any cover. Why is this important? (for that matter, what does it really mean?) While it's easy to stitch together continuous functions to make a continuous function (just sort of join the ends together, right?), it's really hard to stitch smooth functions together in general (join the ends together, and perturb it so the first derivatives align, and so the second align, etc.). But this can be done with little fuss with partitions of unity, and thus with little fuss with second-countability.
And if you study smooth manifolds, you'll see that partitions of unity are immediately used for everything.
So it's the way it is because it has these really nice properties, right? Well, why don't we just require manifolds to be paracompact? (Firstly, there is a distinction, but it's 'small.' A manifold with more than countably many disconnected components may be metrizable, and thus paracompact, but obviously won't be second-countable). In this case, the category of paracompact manifolds is closed coproducts, which doesn't hold for second-countable manifolds. In fact, some people do only consider paracompact manifolds.
At the end of the day, Hausdorff and second countability are exactly what let us use the embedding theorem to view manifolds in real space, and that's what's deemed important for people on their first tour through manifolds.
Let $x\in\partial M$ be any point on the boundary of $M$. Since $M$ is a manifold with boundary, there is an open neighborhood $U$ of $x$ that is homeomorphic to an open subset $V$ of $\mathbb H^n=\{ x\in\mathbb R^n : x_n\ge 0\}$ via a homeomorphism $\phi: U\to V$. Since $x\in\partial M$ we know that $\phi(x)\in V\cap\partial\mathbb H^n$, i.e. $(\phi(x))_n=0$. Since $\phi:U\to V$ is a homeomorphism, it restricts to a homeomorphism
$$\phi^{-1}(V\cap\partial\mathbb H^n) \to V\cap\partial\mathbb H^n.$$
We know that $V\cap\partial\mathbb H^n$ is open in the subspace topology on $\partial\mathbb H^n$ since $V$ is open in $\mathbb H^n$. Thus $V\cap\partial\mathbb H^n$ is an open neighborhood of $\phi(x)$ in $\partial\mathbb H^n$.
Now the open neighborhood $U\cap \partial M$ of $x$ in $\partial M$ is exactly $\phi^{-1}(V\cap\partial\mathbb H^n)$, which as we showed, is homeomorphic to an open subset of $\partial\mathbb H^n\cong\mathbb R^{n-1}$.
On $\partial M$: The most simple definition of $\partial M$ would be the set of all points of $M$ that don't have open neighborhoods which are isomorphic to open sets in $\mathbb R^n$. This gives you that any coordinate chart $\phi:U\to V$ of a manifold with a boundary has to send points of $\partial M$ to points of $\partial \mathbb H^n=\{x\in\mathbb H^n : x_n=0\}$, since all other points in $\mathbb H^n$ have open neighborhoods in $\mathbb R^n$ which can be pulled back to open neighborhoods in $M$ via $\phi$.
Another definition would be that $\partial M$ consists of all points of $M$ that are mapped to points of $\partial\mathbb H^n$ via all (or equivalently one) coordinate chart.
Best Answer
It is false that every second countable space is countable. Second countable means that there exists some countable base $\mathcal{B}$. $\mathbb{R}$ is second countable since a countable number of intervals is a base for $\mathbb{R}$, but it's clearly not countable.
For the forward direction, in your case, since (by definition) $M$ is second countable, and each $U = \{p\}$ is open, and each open set is a union of sets from the basis $\mathcal{B}$, we have that $U \in \mathcal{B}$. Since $\mathcal{B}$ is countable, so is $M$, since $M = \bigcup \mathcal{B}$.
For the backward direction, we have: $M$ is
Hausdorff - consider any two $p,q$ in $M$. Use discreteness of $M$ to conclude immediately.
Second countable (since it's countable)
Locally euclidean. Fix a $p \in M$. Map $p \mapsto 0$ from $U = \{p\} $ to $\{0\}$. Continuous, bijective with countinous inverse.