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 $\sigma$.
Best Answer
Here are some thoughts that could hopefully help.
First of all, trying to give the product of two simplices a simplicial (or $\Delta$-complex) structure is in general annoying. There's a reason that the relevant sections of Hatcher are rather technical (see for example the proof of Theorem 2.10 - page 112 here (PDF)).
Here's one thing that is neat about simplices (when I say simplices I mean $\Delta$-complexes but I'm being sloppy) though, and this construction actually underlies a lot of what you're doing.
Let's suppose that $\sigma = [v_0, \cdots, v_n]$ is an $n$-simplex. Then it has an $(-1)^n$-oriented codimension-1 face $\tau = [v_0, \cdots, v_{n-1}]$, and this is just an $(n-1)$-simplex.
Now let's add a point $v_{n+1}$ to everything - just put it at the end of every simplex. We boost $\sigma$ to an $(n+1)$-simplex $\tilde{\sigma} = [v_0, \cdots, v_n, v_{n+1}]$. And now it has an $(-1)^n$-oriented face $\tilde{\tau} = [v_0, \cdots, v_{n-1}, v_{n+1}]$ which is still codimension-1, and this is just an $n$-simplex.
This is actually super-neat. Let's suppose we have a description of a space $X$ realized as a collection of simplices $\{\sigma_i\}$. Then a simplicial description of $CX$ is given by $\{\sigma_i, [c], \tilde{\sigma}_i\}$, where $\tilde{\sigma}_i$ just denotes adding $c$ to the end of the list of vertices of $\sigma_i$.
If you want to see some poorly-drawn examples, here you go:
Hopefully these convince you that $C(\Delta^n) = \Delta^{n+1}$.
For the suspension, there's a really important point that your notation makes it unclear whether you get it. So I'll say it, and you may well already know it. You identify $Y \times \{0\}$ to one point, and $Y \times \{1\}$ to a different point. Otherwise the equality $\Sigma S^n = S^{n+1}$ wouldn't be true.
Perhaps it's better to think about the suspension as two cones smushed together. Given a space $X$ we can form cones $C_+X$ and $C_-X$, and these come with canonical inclusions $X \to C_{\pm}X$. Then the suspension is just the union of $C_+X$ and $C_-X$ with the two copies of $X$ identified. In the case of the sphere, these correspond to the north and south hemispheres.
In particular, if we start off with $S^0$ as two points (0-simplices), then $S^1 = \Sigma S^0$ is four 1-simplices, one for each quadrant in the standard embedding $S^1 \to \mathbb R^2$, $S^2 = \Sigma S^1$ is eight 2-simplices, one for each octant in $\mathbb R^3$, and so on and so forth.
If you want another poorly-drawn picture, here you go: