$(1)$ The length of the semi-minor axis $b=1$ and the length of the semi-major axis $a=\sqrt{2}$, thus the focal length $c = \sqrt{a^2 - b^2} = \sqrt{2-1}=1$.

Therefore, $F(1,0)$.

Since $l$ is perpendicular to the $x$-axis, its equation is $x=1$.

Solving the system
$ENV1$$$\begin{cases} x=1 \\ \frac{x^2}{2} + y^2 = 1 \end{cases}$$,
we find
$ENV0$$$\begin{cases} x=1 \\ y=\pm \frac{\sqrt{2}}{2} \end{cases}$$.

Thus, $A(1,\frac{\sqrt{2}}{2})$ and $B(1,-\frac{\sqrt{2}}{2})$.

The slopes of lines $AM$ and $BM$ are $\frac{y_B - y_M}{x_B - x_M}=-\frac{\sqrt{2}}{2}$ and $\frac{y_A - y_M}{x_A - x_M}=\frac{\sqrt{2}}{2}$, respectively.

Therefore, the equations of lines $AM$ and $BM$ are $y=-\frac{\sqrt{2}}{2}(x-2)$ and $y=\frac{\sqrt{2}}{2}(x-2)$, respectively.

$(2)$ When $l$ coincides with the $x$-axis, $∠OMA=∠OMB=0^\circ$.

When $l$ is perpendicular to the $x$-axis, $OM$ is the perpendicular bisector of segment $AB$. Therefore, $∠OMA=∠OMB$.

When $l$ neither coincides with nor is perpendicular to the $x$-axis, let the equation of $l$ be $y=k(x-1)$ where $k \neq 0$.

Let $A(x_1,y_1)$ and $B(x_2,y_2)$. Then $x_1 < \sqrt{2}$ and $x_2 < \sqrt{2}$.

The sum of the slopes of lines $MA$ and $MB$ is given by $k_{MA}+k_{MB}=\frac{y_{1}}{x_{1}-2}+\frac{y_{2}}{x_{2}-2}$.

Given $y_{1}=kx_{1}-k$ and $y_{2}=kx_{2}-k$, we have $k_{MA}+k_{MB}= \frac{2k(x_{1}+x_{2})-k(x_{1}x_{2}+2^2)}{(x_{1}-2)(x_{2}-2)}$.

Substituting $y=k(x-1)$ into $\frac{x^2}{2}+y^2=1$ yields $(2k^2+1)x^2-4k^2x+2k^2-2=0$.

Thus, $x_{1}+x_{2} = \frac{4k^2}{2k^2+1}$ and $x_{1}x_{2} = \frac{2k^2-2}{2k^2+1}$.

Therefore, $2kx_{1}x_{2}-3k(x_{1}+x_{2})+4k = \frac{4k^3-6k^3+4k}{2k^2+1} = \frac{-2k^3+4k}{2k^2+1} = 0$.

Hence, $k_{MA}+k_{MB}=0$, indicating that lines $MA$ and $MB$ have complementary angles of inclination.

Thus, $∠OMA=∠OMB$.

In conclusion, \boxed{∠OMA=∠OMB}.