= Solution
Use <integral homology> throughout. The <singular chain group> $C_k(X)$ is free on the <continuous maps> from the standard $k$-simplex to $X$. Since the <boundary operator> preserves chains lying in $A$, these form a <chain subcomplex>. The <relative chain complex> and its <relative homology> are
$$
C_k(X,A)=C_k(X)/C_k(A),\qquad
\boxed{H_k(X,A)=\ker(\partial:C_k(X,A)\to C_{k-1}(X,A))/\operatorname{im}(\partial:C_{k+1}(X,A)\to C_k(X,A)).}
$$
The <short exact sequence of chain complexes> gives the <long exact sequence in relative homology>.
A sufficient hypothesis for the quotient comparison is a nonempty <good pair>: $A$ is closed and has an open neighbourhood $U$ which admits a <deformation retraction> onto $A$ while fixing $A$ throughout. A <CW pair> is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular <collar neighbourhood>. The induced <quotient map> $q:X\to Y=X/A$ collapses $A$ to a point $a$.
Here is the proof of the <collapsing a pair theorem>. Put $V=U/A\subset Y$. The <deformation retraction> makes $H_*(U,A)=0$, so the <long exact sequence> of the triple gives $H_*(X,A)\cong H_*(X,U)$. It also contracts $V$ onto $a$, giving $H_*(Y,\{a\})\cong H_*(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, <excision> gives
$$
H_*(X,U)\cong H_*(X\setminus A,U\setminus A).
$$
Similarly, <excision> of the closed point $a$ in the open set $V$ gives
$$
H_*(Y,V)\cong H_*(Y\setminus\{a\},V\setminus\{a\}).
$$
The <quotient map> restricts to a <homeomorphism> between the two punctured pairs, so these are the same groups and the identifications commute with $q_*$. Finally $H_*(Y,\{a\})\cong\widetilde H_*(Y)$, including degree zero. Thus
$$
\boxed{q_*:H_k(X,A)\xrightarrow{\sim}\widetilde H_k(X/A).}
$$
The neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.
For the genus-two <closed orientable surface> with its chosen <orientation>, $H_0(X)=\mathbb Z$, $H_1(X)=\mathbb Z^4$, $H_2(X)=\mathbb Z$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb Z$. Its relevant <long exact sequence in relative homology> is
$$
0\longrightarrow\mathbb Z\longrightarrow H_2(X,A)
\longrightarrow\mathbb Z\xrightarrow{i_*}\mathbb Z^4
\longrightarrow H_1(X,A)\longrightarrow\mathbb Z\xrightarrow{\sim}\mathbb Z\longrightarrow0.
$$
The final map is an <isomorphism> because both spaces are connected. Consequently $H_1(X,A)=\operatorname{coker}i_*$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The <quotient topological space> is connected, so its unreduced $H_0$ is $\mathbb Z$.
In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its <homology class> in $X$ is therefore zero, so $i_*=0$. The <exact sequence> becomes
$$
0\to\mathbb Z\to H_2(X,A)\to\mathbb Z\to0,
\qquad H_1(X,A)=\mathbb Z^4.
$$
The first sequence splits since its quotient is a free <abelian group>. Hence the <homology after collapsing a separating surface curve> is
$$
\boxed{H_k(X/A;\mathbb Z)\cong\begin{cases}
\mathbb Z,&k=0,\\
\mathbb Z^4,&k=1,\\
\mathbb Z^2,&k=2,\\
0,&k\geq3.
\end{cases}}
$$
Geometrically, collapsing the boundary of each once-punctured <torus> fills its puncture with a cone on the <circle>, which is a disk. The quotient is a <wedge sum> of two tori at the collapsed point. Their two independent <fundamental classes> explain the extra second-homology generator.
In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed <intersection pairing on an oriented surface> therefore provides an integer homomorphism $H_1(X)\to\mathbb Z$ taking $[A]$ to $\pm1$. Thus $[A]$ is a nonzero <primitive homology class>, and $i_*:\mathbb Z\to\mathbb Z^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is $\mathbb Z^3$. The same <exact sequence> gives $H_2(X,A)\cong H_2(X)\cong\mathbb Z$. Therefore the <homology after collapsing a nonseparating surface curve> is
$$
\boxed{H_k(X/A;\mathbb Z)\cong\begin{cases}
\mathbb Z,&k=0,2,\\
\mathbb Z^3,&k=1,\\
0,&k\geq3.
\end{cases}}
$$
Another description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed <torus> with two marked points, then identify the two points. Identifying two distinct points in a connected <CW complex> adds a loop up to <homotopy>, so this quotient has the <homotopy type> of $T^2\vee S^1$. The preceding <relative homology> calculation proves the groups without requiring that <homotopy> description.
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