For the case of a closed oriented surface $S$, its first Betti number $\beta_1$ and its genus $g$ are related by the equation
$$\beta_1 = 2g
$$
Those two quoted phrases therefore cannot be the same. Notice that the cut procedure in the phrase defining genus is quite precise, whereas the cut procedure defining Betti number is rather vague, and therein lies the difference. You might want to follow up the reference given in the Betti number definition, which occurs right after your quote cuts off. One traditional way to describe $\beta_1$ is as the maximal number $2k$ of simple closed curves in $S$ that can be listed as $a_1, b_1, a_2, b_2,...,a_k,b_k$ so that if $i \ne j$ then $a_i$ and $b_i$ are disjoint from $a_j$ and $b_j$, and so that $a_i,b_i$ intersect each other transversely exactly one time.
Your question also raises several other issues, and I'll treat these briefly.
The equation $\beta_1 = 2g$ that relates the first betti number and the genus can be deduced by comparing the actual definition of $\beta_1$, namely the rank of the first homology group, with the actual calculation of the first homology group of the surface (carried out by using any of the calculational procedures learned in algebraic topology). I don't think it's particularly fruitful to over-interpret the meaning of an intuitive explanation of betti number in terms of "number of holes". The purpose of that intuition is to get you started, and you could keep reading other posts on that issue if you desire, but really you should dig into the technical details of homology groups to learn what's really going on.
In particular, the relation between the Euler characteristic and the genus can again be deduced by working through the calculations of homology groups.
If you want more detail, it would be better to ask separate and more precisely worded questions, instead of lumping too many vaguely worded questions into one post.
There does always exist such a surface. There is a reasonably short proof using a big tool, namely Selberg's Lemma that every linear group has a torsion free subgroup of finite index.
Assuming equilateral triangles, the symmetry group of the tiling of the hyperbolic plane $\mathbb H^2$ that you depict is a reflection group. In particular, if you pick any tile, then subdivide it into 6 triangles each with angles $\pi/2$, $\pi/3$, $\pi/n$, and then take $T$ to be any one of those 6 triangles, the symmetry group is generated by reflections in the three sides of $T$. That group is a Coxeter group
$$C(2,3,n) = \langle a,b,c \mid a^2 = b^2 = c^2 = (ab)^2 = (bc)^3 = (ca)^n = \text{Id} \rangle
$$
Furthermore, $T$ is a fundamental domain for the action of $C(2,3,n)$. These conclusions all come from the Poincare Fundamental Polygon Theorem.
We can pass to an index 2 subgroup $R(2,3,n) < C(2,3,n)$ in which the sum of the exponents of $a,b,c$ in each word is even. One can show that $R(2,3,n)$ is generated by the rotations $ab$, $bc$ and $ca$ of order $2,3,n$ respectively, with defining relation $(ab)(bc)(ca) = \text{Id}$. The action of $R(2,3,n)$ on $\mathbb H^2$ therefore preserves orientation. The orientation preserving isometry group of the hyperbolic plane is a linear group, namely $SO(2,1)$. Selberg's Lemma therefore applies, yielding a torsion free finite index subgroup $\Gamma < R(2,3,n) < C(2,3,n)$. The quotient $\mathbb H^2 / \Gamma$ is the surface that you desire.
Best Answer
The connected sum of two disks is an annulus. If you think of an annulus as being a hole, then I suppose a disk is half a hole.