You cannot make both assumptions about \$V_{ce} = 0.2V\$ and \$\beta = 100\$ at the same time; they are for two different modes of operation for the transistor (there's also a third mode where the transistor is in cutoff, and a fourth less common one in reverse active).
You start out assuming one of the possible states. For example, suppose I assume the transistor is saturated. Then \$V_{ce} = 0.2V\$ and \$V_{be} = 0.7V\$. You can then solve for base and collector currents using these assumptions for the BJT and only these assumptions.
The last step is to check your assumptions. For example, to make sure the transistor isn't actually in the linear active region must check that \$\beta I_{b} \gg I_{c}\$. So let's say in your case you get \$I_{c} = 9.8 mA\$ and \$I_{b} = 9.8 \mu A\$. Well clearly our assumption about saturation was bad because \$\beta I_{b} = 980 \mu A\$, which is less than our calculated \$I_{c}\$. We must then start over with new assumptions and re-solve the problem.
Instead, suppose \$I_b = 2 mA\$. Then \$\beta I_b = 200mA\$, which is much greater than \$I_c\$, so now the saturation assumption is correct.
Note: \$\beta I_b \gg I_c\$ is kind of a vague limit. Typically we use at least 10 times bigger for saturation.
For solving assuming the BJT is forward active, you would assume \$\beta I_b = I_c\$ and only this assumption. To check it (against saturation), simply make sure \$0.2V < V_{ce}\$.
One way to understand a transistor when it is used as a switch is to assume that when the transistor is fully "on" the voltage between its emitter and base is 0.6V.
If you look at your system, when the Arduino sets the PWM pin to 5V. the voltage between the base and the emitter is 0.6V. Thus the voltage across the motor is 5-0.6=4.4V.
I don't know what's the specs of your motor, but 4.4V seems too small.
Using your components, there is a way to obtain almost 9V across your motor.
See the following schematics :
simulate this circuit – Schematic created using CircuitLab
Here you can see that when your transistor is conducting, the motor would see almost 9V and may run faster than just "buzzing".
Best Answer
You really should use component designators in your schematics. Lack of them makes it more difficult to talk about the circuit, so I'll be more brief.
Consider what happens when the reed switch is open. The first transistor is on, lighting the first LED. The collector being low also turns off the second transistor, which keeps the second LED off.
When the reed switch is closed, the first transistor is forced off, so the first LED will be off too. The second transistor is then turned on, which lights the second LED.
In summary, one LED is always lit. The reed switch state governs which one.