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 $\sigma$.
Some context: the question refers to Hatcher's proof of the Lefschetz fixed point theorem. The proof goes like this: Given a finite simplicial complex $X$, he wants to show that if some map $f:X \to X$ has no fixed points, then $\tau(f_*) = 0$. He does this by constructing a map $g$, homotopic to $f$, which is simplicial when considered as a map from some subdivision $K$ of $X$ to itself, such that $g(\sigma) \cap \sigma = \emptyset$ for all simplices $\sigma$ in this subdivision. He further shows that the Lefschetz number $\tau(f)$ can be computed by
$$
\tau(f) = \tau(g) = \sum (-1)^n \text{tr} \left( g_*: H_n(K_n, K_{n-1}) \to H_n(K_n, K_{n-1})\right)
$$
Then he claims that each summand on the right side is zero. This is what the question is asking about. The group $H_n(K_n, K_{n-1})$ is free abelian, with basis the $n$-simplices of $K$, and the map $g_*: H_n(K_n, K_{n-1}) \to H_n(K_n, K_{n-1})$ will take the generator for an $n$-simplex $\sigma$ to (up to a sign) the generator for the simplex $g(\sigma)$, or to zero if $g(\sigma)$ is not $n$-dimensional. So since $g(\sigma) \cap \sigma$ is empty, the matrix for $g_*$ has zeroes down the diagonal.
Best Answer
Hint: by definition, a simplicial map $f : K \to L$, maps the vertices of an $i$-simplex $\sigma$ of $K$ onto a set of vertices in $L$ that span a $j$-simplex $\tau$ of $L$, where necessarily $j \le i$. So $f$ maps the $i$-skeleton of $K$ (which comprises the union of the $k$-simplices in $K$ with $k \le i$) into a union of $l$-simplices in $L$ with $l \le i$, which is contained in the $i$-skeleton of $L$ (which comprises the union of all the $l$-simplices in $L$ with $l \le i$).