First, it's not clear to me how to prove (assuming it's true, as in the smooth case) that the fact that $f$ is orientation preserving at a point $x \in M$ implies that it is orientation preserving at every point. That is, I'd like to prove that if for some $x$ in $M$ and $f_*: H_n(M, M - \{ x\}) \to H_n(M, M - \{ f(x)\}) $ we have that $f_*([M]_x) = [M]_{f(x)}$, then the same holds for every point in $M$ (here $[M]_x$ is our selected generator for $H_n(M, M-\{ x\})$).
This is true as long as $M$ is connected. For any $x\in M$, there is some open set $U$ containing $x$ and an element $[M]_U\in H_n(M,M-U)$ such that for all $y\in U$, the image of $[M]_U$ in $H_n(M,M-\{y\})$ is the generator $[M]_y$ chosen by our orientation. Shrinking $U$, we may assume that there is also such a class $[M]_{f(U)}\in H_n(M,M-f(U))$. We may further shrink $U$ to assume $U$ is an open ball in some neighborhood of $x$ homeomorphic to $\mathbb{R}^n$ so that the map $H_n(M,M-U)\to H_n(M,M-\{y\})$ is an isomorphism for each $y\in U$ (and similarly for $f(U)$) and so the elements $[M]_U$ and $[M]_{f(U)}$ are unique.
We then have that for any $y\in U$, $f$ sends $[M]_y$ to $[M]_{f(y)}$ iff $f_*:H_n(M,M-U)\to H_n(M,M-f(U))$ sends $[M]_U$ to $[M]_{f(U)}$. The latter condition does not depend on the point $y\in U$. Thus if $f$ is orientation-preserving at $x$, it is orientation-preserving in a neighborhood of $x$, and similarly if it is orientation-reversing. So the sets of points where $f$ is orientation-preserving and where $f$ is orientation-reversing are an open partition of $M$. If $M$ is connected, one of these sets must be all of $M$.
Second, I'd have to show that $f_*([M]_x) = [M]_{f(x)}$ for some $x$ implies $g_*([M]_x) = [M]_{g(x)}$.
It suffices to show that $f_*([M]_x) = [M]_{f(x)}$ implies $(f_t)_*([M]_x) = [M]_{f_t(x)}$ for all sufficiently small $t$. Fixing a neighborhood $V\cong\mathbb{R}^n$ of $f(x)$, there is an open neighborhood $U$ of $x$ and an $\epsilon>0$ such that $f_t(U)\subseteq V$ for all $t\in[0,\epsilon]$. We may as well assume $\epsilon=1$, $U$ is our entire domain, and $V$ is our entire codomain. That is, we may as well assume we have a homotopy consisting of open embeddings $f_t:U\to\mathbb{R}^n$ for each $t\in [0,1]$, and wish to show $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$.
But now we can modify our maps $f_t$ so that we can look at just one point: let $f_t'(y)=f_t(y)-f_t(x)$. Note that $f_t'$ is orientation-preserving at $x$ iff $f_t$ is orientation-preserving at $x$, since it differs by composition with a translation, which is orientation-preserving on $\mathbb{R}^n$. We also have $f_t'(x)=0$ for all $t$. So, by considering the induced maps $H_n(U,U-\{x\})\to H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})$, we see that $f_0'$ is orientation-preserving at $x$ iff $f_1'$ is orientation-preserving at $x$. It follows that $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$, as desired.
[In this argument I am using the fact that $\mathbb{R}^n$ has only two different orientations, so that when we orient the codomain $\mathbb{R}^n$ in the same way as our given orientation on $V$, we have either the standard orientation or its opposite. This guarantees that translation is orientation-preserving on $\mathbb{R}^n$, for our chosen orientation. The fact that $\mathbb{R}^n$ only has two orientations is because it is connected, using the first part of the answer above.]
