Direction of path depends on sign of determinant of Jacobian Let $f: K_1(0) \rightarrow \mathbb{R}^2$ be continuously differentiable, $\{z_1,..,z_m\}=f^{-1}(a)$ with a regular $a \in \mathbb{R}^2$.
We choose $\epsilon$ small enough so that $f\vert_{\overline{U}_\epsilon(z_i)}$ are homeomorphisms.
Consider $\partial U_\epsilon(z_i)=:S_i$. Also view $S_i$ as the path surrounding the $\epsilon$-circle in the mathematically positive direction.
 Then $f(S_i)$ is a Jordan curve with $a$ on the inside of $f(S_i)$, since $f$ is a homeomorphism on $S_i$.
$f(S_i)$ has the same orientation as $S_i$, if $det ~J_f(z_i) > 0$ and the inverse orientation, if $det ~J_f(z_i)<0$.
Why does the determinant of the Jacobian decide the orientation of $f(S_i)$?
 A: We shall use the fact that the winding number is homotopy invariant.
We only consider $S = S_1$. Let $L : \mathbb{R}^2 \to \mathbb{R}^2$ denote the linear isomorphism of $\mathbb{R}^2$ corresponding to $J_f(z_1)$; it is characterized by $\lim_{u \to 0}\frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} = 0$. Let $r = \min\{ \lVert L(z)\rVert \mid \lVert z \rVert = 1 \}$. We have $r > 0$ because $L$ is an isomorphism. Then
$$\frac{\lVert f(z_1 + u) - f(z_1) \rVert}{\lVert u \rVert} \ge \frac{\lVert L(u) \rVert}{\lVert u \rVert} - \frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} = \\
\lVert L(\frac{u}{\lVert u \rVert}) \rVert- \frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} > \frac{r}{2}$$
for $0 < \lVert u \rVert \le \mu$ (we may of course assume that $\mu \le \epsilon_1$). This shows that there exists $0 < \rho \le \mu$ such that
$$(\ast) \phantom{xx} \lVert f(z_1 + u) - f(z_1) - L(u) \rVert < \lVert f(z_1 + u) - f(z_1) \rVert$$
for $0 < \lVert u \rVert \le \rho$ because otherwise we would obtain a sequence $(u_n)$ such that  $0 < \lVert u_n \rVert \le \mu$ and $u_n  \to 0$ and
$$\frac{\lVert f(z_1 + u_n) - f(z_1) - L(u_n) \rVert}{\lVert u_n \rVert} \ge \frac{\lVert f(z_1 + u_n) - f(z_1) \rVert}{\lVert u_n \rVert} > \frac{r}{2} .$$
A geometric consequence of $(\ast)$ is
$$(\ast\ast) \phantom{xx} \text{The line segment connecting } f(z_1 + u) \text{ and } f(z_1) + L(u) \text{ does not contain } \\
a = f(z_1)$$
for $0 < \lVert u \rVert \le \rho$.
Now let $u : [0,1] \to K_1(0)$ be the closed path $u(t) = z_1 + \epsilon_1e^{2\pi i t}$ (here $\mathbb{C}$ enters again, but only for convenient notation). It is homotopic in $K_1(0) \backslash \{ z_1 \}$ to the path $v(t) = z_1 + \rho e^{2\pi i t}$. Hence $f \circ u$ and $f \circ v$ are homotopic in $\mathbb{R}^2 \backslash \{ a \}$ and therefore have the same winding number. We claim that the paths $f \circ v$ and $w(t) = a + L(\rho e^{2\pi i t})$ are homotopic in $\mathbb{R}^2\backslash \{ a \}$. Define $H : [0,1] \times [0,1] \to \mathbb{R}^2\backslash \{ a \}, H(t,s) = (1-s)f(v(t)) + sw(t)$. By $(\ast \ast)$ this is well-defined (note $f(v(t)) =  f(z_1 + \rho e^{2\pi i t}), w(t) = f(z_1) + L(\rho e^{2\pi i t})$). We see that $H_0 = f \circ v, H_1 = w$ and all $H_t$ are closed paths  (i.e. $H_t(0) = H_t(1)$).
It therefore suffices to determine the winding number of $w$ which can be safely left as an exercise. Note that the winding number of $w$ around $a$ is the same as the winding number of $\omega(t) = L(\rho e^{2\pi i t})$ around $0$ and the latter has the same winding number as $\sigma(t) = L(e^{2\pi i t})$ (use again a homotopy).
The "philosophy" behind this proof is this: The differentiable maps are precisely those which can be nicely approximated by affine maps around each  point. This implies that each sufficiently small circle around a point in which $f$ has a non-singular Jacobian is mapped to a Jordan curve which can be nicely approximated by the image of the circle under an affine map whose linear part is the derivative of $f$.
Remark concerning the winding number of the path $\sigma$:
The corresponding integral can be computed explicitly although it is somewhat tedious and requires some knowledge about the integration of trigonometric expressions of the form $\frac{1}{a \cdot sin^2t + b \cdot sint \cdot cost + c \cdot cos^2t}$. You will find that the the winding number of $\sigma$ is  $\frac{detL}{\lvert detL \rvert}$.
An alternative approach is to use once more the homotopy invariance of the winding number and to construct a homotopy from $\sigma$ to one of the paths $a(t) = e^{2 \pi it }$ (if $detL > 0$) or $b(t) = e^{-2 \pi it }$ (if $detL < 0$).
Let $e_1 = (1,0), e_2 = (0,1)$ and let $R(\alpha)$ denote the rotation by the angle $\alpha$. There exists $\varphi$ such that $R(\varphi)(L(e_1)) = re_1$ with $r = \lVert L(e_1) \rVert > 0$. Then $\sigma$ is homotopic to $\sigma_1(t) = (R(\varphi)L)(e^{2\pi i t})$ (via $h(t,s) = (R(s\varphi)L)(e^{2\pi i t})$). Rotating is essential to avoid that $h(t,s) = 0$ for some $s$. $L_1 = R(\varphi)L$ is a linear isomorphism having the same determinant as $L$. We have $L_1(e_1) = re_1 = (r,0)$ and $L_1(e_2) = (w_1,w_2)$; this shows $det L = det L_1 = rw_2$. Let $\epsilon = sign(det L)$. Then $\sigma_1$ is homotopic to $a_\epsilon(t) = e^{\epsilon 2 \pi it }$ (via $g(t,s) = (1-s)\sigma_1(t) + sa_\epsilon(t)$; note that $g(t,s) \ne 0$ for all $t,s$).
