The number of patients treated is cumulative, so the outcome of interest here is that the doctor has treated three additional patients between $8$ and $10$ a.m.
One of the properties of the Poisson process is that, at any given point in time, the occurrence of future events is independent of any past events. So all you have to do is compute the probability that the number of arrivals $X(t)$ in the next $t = 2$ hours equals $3$, when the rate is $\lambda = 2$ arrivals per hour. This is given by $$\Pr[X(2) = 3] = e^{-\lambda t} \frac{(\lambda t)^3}{3!} = \frac{32}{3e^{4}} \approx 0.195367.$$
But suppose we don't do it this way, and try to model the problem more along the lines of what you were thinking. Then let $A$ be the event that exactly $6$ arrivals occurred by time $t = 2$ hours, and let $B$ be the event that exactly $9$ arrivals have occurred by time $t = 4$ hours. Then $$\Pr[B \mid A] = \frac{\Pr[B \cap A]}{\Pr[A]}.$$ The denominator is easy; with $\lambda t = 2(2) = 4$, we have $$\Pr[A] = e^{-4} \frac{4^6}{6!} \approx 0.104196.$$ This is the same as what you computed. Where we differ is the numerator; in particular, $$\Pr[B \cap A] \ne \Pr[B].$$ The reason is because the event $B \cap A$ excludes outcomes in which, say, only $5$ arrivals occurred in the first two hours but $4$ occurred in the next two hours; whereas $\Pr[B]$ considers whether a total of $9$ arrivals occurred within the first four hours irrespective of their distribution within the first and second halves of the four-hour interval. That's why your probability exceeds $1$: your numerator becomes too large because $\Pr[B]$ includes events you should be excluding.
This leads us to conclude that the intuitive way to compute $\Pr[B \cap A]$ is to reason that $\Pr[B \cap A] = \Pr[B \mid A]\Pr[A]$ and $\Pr[B \mid A] = \Pr[X(2)=3]$, which is of course circular because what we were after from the beginning was $\Pr[B \mid A]$. So is there some other way? Well, yes, but it's not simple. The idea is to consider the distribution of the $9$ arrivals in the first four hours according to whether they happen in the first two or last two hour intervals. In other words, we think of how many ways they can be distributed between the two intervals, and count only the one where $6$ happen in the first half and $3$ in the second.
Because the intervals have equal length, given that an arrival occurred in the first four hours, the probability that it occurs in the first half equals the probability it occurs in the second. So the probability that exactly $Y = 6$ events occurred in the first half is a binomial probability with $n = 9$ and $p = 1/2$: $$\Pr[Y = 6] = \binom{9}{6} (1/2)^6 (1 - 1/2)^3 = \frac{21}{128} \approx 0.164063.$$ Then we have to also multiply this by the Poisson probability of seeing $9$ events in the first place, i.e. $$Pr[B \cap A] = \Pr[Y = 6]\Pr[B] \approx 0.164063(0.124077) \approx 0.0203564.$$ And when we finish the calculation, you find $$\Pr[B \mid A] = \frac{0.0203564}{0.104196} \approx 0.195367,$$ which matches our original calculation at the beginning.
What the equation $\Pr[B \cap A] = \Pr[Y = 6]\Pr[B]$ suggests is that in fact, $$\Pr[Y = 6] = \frac{\Pr[B \cap A]}{\Pr[B]} = \Pr[A \mid B].$$ This is the conditional probability that, if $9$ events were observed in the first $4$ hours, that $6$ of them happened in the first two hours. And this is indeed binomially distributed, but we did not really formally prove this, resorting only to intuition.
As clarified in the comments above, we seek to find the average number of patients in a randomly-selected vehicle arriving at the hospital. As is correctly stated in the question, such a vehicle will either be from District 1 or District 2 with average numbers of patients $\lambda/\theta$ and $\gamma$, respectively.
The probabilities (i.e. the weights) for finding a vehicle from each District are simply the mean dispatch frequencies from the two districts, normalized to sum to 1:
$$
P(\text{Vehicle is from District 1}) \;=\; \frac{\theta}{\theta + (\lambda/\gamma)}
$$
$$
P(\text{Vehicle is from District 2}) \;=\; \frac{\lambda/\gamma}{\theta + (\lambda/\gamma)}
$$
The distances of each district from the hospital have no bearing, since - assuming the system is at equilibrium - every vehicle dispatched to the hospital arrives at the hospital, so the pipeline to the hospital will be continuously populated by previously-dispatched vehicles.
Another way to look at it: If the arrival frequency were greater than the dispatch frequency, it would mean extra vehicles were being conjured into existence en route. If the arrival frequency were less than the dispatch frequency, it would mean vehicles were building up in the pipeline.
Best Answer
The main theorem you can use is the one that links Poisson and Negative Exponential (it is very easy to be proved).
Thus any event represented with a negative exponential (i.e. lifetime of some devices) but viewed as # of arrival can be represented by a Poisson