Yes, this is true. Here's one way to prove it. For $0\leq j<n$, write $s_j^{n-1}:\Delta^n\to\Delta^{n-1}$ for the map induced by the order-preserving surjection $\{0,\dots,n\}\to\{0,\dots,n-1\}$ that maps both $j$ and $j+1$ to $j$. Say that a singular simplex $\Delta^n\to X$ is degenerate if it factors through $s_j^{n-1}$ for some $j$. Note that the boundary of a degenerate simplex is a linear combination of degenerate simplices: all but possibly two of its faces are degenerate (the two exceptions being the faces corresponding to omitting the vertices $j$ and $j+1$), and those two faces cancel out. So if we write $D_n(X)\subseteq C_n(X)$ for the span of the degenerate simplices, $D_\bullet(X)$ is a subcomplex of $C_\bullet(X)$. Write $N_\bullet(X)=C_\bullet(X)/D_\bullet(X)$.
Now note that given an order-preserving simplicial map $f:X\to Y$, the induced maps $C^\Delta_\bullet(X)\to C^\Delta_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)$ and $C^\Delta_\bullet(X)\to C_\bullet(X)\to C_\bullet(Y)\to N_\bullet(Y)$ are equal, since the $n$-simplices $\sigma$ that you're worried about have the property that $f\circ c_\sigma$ is degenerate. So to show that $f^\Delta_{\bullet_*}=f_{\bullet_*}$, it suffices to show that the map $C_\bullet(Y)\to N_\bullet(Y)$ induces isomorphisms on homology. By the long exact sequence in homology associated to the short exact sequence $0\to D_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)\to 0$, it suffices to show that $D_\bullet(Y)$ has trivial homology.
We can show this by constructing a chain homotopy. Given a degenerate $n$-simplex $\sigma:\Delta^n\to Y$, let $j(\sigma)\in\{0,\dots,n-1\}$ be the least $j$ such that $\sigma$ factors through $s_j^{n-1}$. Now define $H:D_n(Y)\to D_{n+1}(Y)$ by $H(\sigma)=(-1)^{j(\sigma)}\sigma\circ s^n_{j(\sigma)}$. An elementary computation then shows that $H\partial+\partial H:D_n(Y)\to D_n(Y)$ is the identity map. It follows that $D_\bullet(Y)$ has no homology.
To give a sense of the computation $H\partial+\partial H=1$, let's show what happens when you take a $3$-simplex $\sigma$ with $j(\sigma)=1$; the general case works very similarly. Let's write $\sigma=[a,b,b,c]$; all the simplices built from $\sigma$ will be written using similar expressions with the obvious meaning (here $a$, $b$, and $c$ are the vertices of $\sigma$, with $b$ repeated since it factors through $s^2_1$). We then have $H(\sigma)=-[a,b,b,b,c]$, so $$\begin{align}\partial H(\sigma)=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,c]+[a,b,b,c]-[a,b,b,b]\\=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,b].\end{align}$$
On the other hand, $\partial\sigma=[b,b,c]-[a,b,c]+[a,b,c]-[a,b,b]=[b,b,c]-[a,b,b]$. We have $H([b,b,c])=[b,b,b,c]$ and $H([a,b,b])=-[a,b,b,b]$. Thus we get $$H(\partial\sigma)=[b,b,b,c]+[a,b,b,b].$$ When we add together $\partial H(\sigma)$ and $H(\partial\sigma)$, all the terms cancel except $[a,b,b,c]$, which is just $\sigma$.
You excise the set $B = S^n - \bigcup_{i=1}^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^{-1}(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^{-1}(y)) \approx H_n(\bigcup_{i=1}^k U_i, \bigcup_{i=1}^k (U_i - x_i)) = H_n(\bigcup_{i=1}^k( U_i, U_i - x_i)) \approx \bigoplus_{i=1}^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^{-1}(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^{-1}(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
Best Answer
For $n \ne k$ both of $H^\Delta_n(X^k,X^{k-1})$ and $H_n(X^k,X^{k-1})$ are trivial groups, and any homomorphism from one trivial group to another is an isomorphism.
For $n = k$, both of $H^\Delta_n(X^k,X^{k-1})$ and $H_n(X^k,X^{k-1})$ are free abelian groups, with specific bases for each. Also, the homomorphism $H^\Delta_n(X^k,X^{k-1}) \mapsto H_n(X^k,X^{k-1})$ takes the basis of the first bijectively to the basis of the second. Any homomorphism from one free abelian group to another, that takes a basis of the first to a basis of the second bijectively, is an isomorphism.