Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn't be an answer...)
First some preliminary notions:
For a topological space $X$, an $n$-simplex in $X$ is a continuous map $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.
An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th vertex of $n_+$.
Then for the two homologies:
The singular (unreduced) chain complex on a space $X$, is the chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0$$
where $C_n(X)$ is the free abelian group $\mathbb{Z}[S_n(X)]$ generated by the set $S_n(X) = \{ \sigma : \Delta^n \to X \}$ of all $n$-simplices in $X$ (i.e. the set of all continuous maps $\Delta^n \to X$). The boundary maps $\partial_n : C_n(X) \to C_{n-1}(X)$ is given by $\partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X$.
The $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th singular homology group of $X$.
A simplicial complex $S$ is a set $S = \bigcup_{n=0}^{\infty} S_n$ where $S_n = S(n_+)$ being a set of ordered $n$-simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. The simplicial chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0$$
consists of the free abelian groups $C_n(S) = \mathbb{Z}[S_n]$ generated by the $n$-simplices. The boundary map $\partial_n : C_n(S) \to C_{n-1}(S)$ is given by $\partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma$ where $d_i = S(d^i) : S_n \to S_{n-1}$ is the face maps $d_i(\sigma) = \sigma \circ d^i$.
The $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th simplicial homology group of $S$.
Lastly we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$, where $d_i \sigma \times \Delta^{n-1}$ is identified with the $i$'th face of $\sigma \times \Delta^n$.
Then if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex).
NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..
There are relative simplicial homology groups $H^\Delta_*(X,A)$ where $A$ is a subcomplex of a $Δ$-complex. These are the homology groups of the chain complex in the third row of the following commutative diagram, where $Δ_n(X,A)$ is the quotient $Δ_n(X)/Δ_n(A)$, so the columns are exact. This also gives rise to a long exact sequence of simplicial homology groups as in the singular case.
$\qquad$
$$0\to H^Δ_2(M,∂M)→H^Δ_1(∂M)→H^Δ_1(M)→H^Δ_1(M,∂M)→\tilde H^Δ_0(∂M)=0$$
The key is to compute the map $H^Δ_1(∂M)\to H^Δ_1(M)$. Looking at the $Δ$-complex structure as in the picture on the right, we see that $H^Δ_1(M)$ is generated by $c$, and $H^Δ_1(∂M)$ is generated by $a+a'$, which in $H^\Delta_1(M)$ is homologous to $(b+a')+(a-b)\sim c+c$. That means the map is multiplication by $2$, and this determines the relative groups $H^Δ(M,∂M)$, using exactness of the sequence.
Edit: I just see that your question was about equivalence of singular and simplicial relative homology groups. This is Theorem 2.27 in Hatcher's Algebraic Topology.
Best Answer
Let $S^{n-1}$ be the complex with $\mathbb{Z}$ in degree $0, n-1$ and $0$ in all other degrees. Let $i_n$ be the unique nondegenerate $n$-simplex of $\Delta^n$. Then the nondegenerate $(n-1)$-simplices of $\partial\Delta^n$ are $d^n_i(i_n), i=0,\ldots, n$ and $c = \sum_{j=0}^n(-1)^jd_j(i_n)$ is a cycle in $N(\partial \Delta^n)$, since it is a boundary in $N(\Delta^n)$. Define a chain map $f\colon S^{n-1}\to N(\partial\Delta^n)$ via $f_{n-1}(k) = kc$ and in degree $0$ $\mathbb{Z}\to N(\partial\Delta^n)_0$ by sending the generator of $\mathbb{Z}$ to the basis element corresponding to the $0$th vertex. There is a retraction $r\colon N(\partial\Delta^n)\to S^{n-1}$. In degree $0$ this maps $N(\partial\Delta^n)_0\to \mathbb{Z}$ the basis element corresponding to the $0$th vertex to $1$ and all other basis elements to $0$. In degree $n-1$ it is given by $r_{n-1}\colon N(\partial\Delta^n)\to \mathbb{Z}$, $$r_{n-1}(d^n_j(i_n)) = \begin{cases} 0 &\text{if }j<n,\\ 1 &\text{if }j=n. \end{cases}$$ I claim that $f\circ r$ is homotopic to the identity. An increasing map $\alpha\colon [k]\to [n]$ corresponds to an increasing sequence $(i_0,\ldots, i_k)$ and this is nondegenerate if all elements are distinct. We define a chain homotopy on basis elements by $$s_k(i_0,\ldots,i_k) = \begin{cases} (-1)^{k+1}(i_0,\ldots, i_k, n) &\text{if } i_k<n,\\ 0 &\text{otherwise}, \end{cases}$$ and we extend this to a linear map $N(\partial\Delta^n)_k\to N(\partial\Delta^n)_{k+1}$. I leave it to you to check that this gives the desired chain homotopy.
One can also use the long exact sequence in homology. Let $\mathbb{Z}[n]$ be the complex with $\mathbb{Z}$ in degree $n$ and $0$ in all other degrees. Then there is a short exact sequence of complexes $$0\to N(\partial\Delta^n)\to N(\Delta^n)\to \mathbb{Z}[n]\to 0,$$ since $\partial\Delta^n$ and $\Delta^n$ have the same nondegenerate $k$-simplices for $k\neq n$, $\partial\Delta^n$ has only degenerate $n$-simplices and $\Delta^n$ has exactly one nondegenerate $n$-simplex. Now if you already know the homology of $N(\Delta^n)$ it is easy to read off the homology of $N(\partial \Delta^n)$ from the long exact sequence in homology.