My question now: considering I use the 10th percentiles of those two input variables, and run these values through my model. Is my output now also the 10th percentile of its own probability distribution (which is a combination of the two distributions of the inputs I think)?
No this is not generally true.
A counter example is the addition of two standard normal distributed variables
$$Z = X + Y \quad \text{where X,Y i.i.d. $\mathcal{N}$(0,1)}$$
Then we do not have that the quantile function follow the same formula
$$Q_Z(p) \neq Q_X(p) + Q_Y(p)$$
Instead,
$$Q_Z(p) = \frac{ Q_X(p) + Q_Y(p) }{\sqrt{2}}$$
(You can derive this by considering that the quantile function is a constant multiplied by the standard deviation, e.g. the 0.975-th quantile is 1.96 times the standard deviation. But $\sigma_Z = \sqrt{2}$ and $\sigma_X = \sigma_Z = 1$, so $\sigma_Z \neq \sigma_X+\sigma_Y$ and instead $\sigma_Z =(\sigma_X+\sigma_Y)/\sqrt{2}$)
When does it work?
The case when this idea works is when you think about a monotonic transformation of a variable $$Z = f(X)$$
where monotonic means $f(x_1) \leq f(x_2)$ if and only if $x_1 \leq x_2$.
Then the cumulative distribution functions have the following correspondence $$P[Z \leq f(a)] = P[f(X) \leq f(a)] = P[X \leq a] $$and the quantile functions will be related as $$Q_Z(p) = Q_{f(X)}(p) = f(Q_X(p))$$
Methods to compute the distribution of the output
My current thought: it is impossible to determine this analytically (even in the case of assumed normality of the inputs), and it could be useful to apply simulation. That is, generate a thousand samples from the probability distributions of the two inputs (assuming they are independent) and run those samples through the model in order to find the approximated 10th percentile from the probability distribution of my output variable.
For some cases, you can derive the distribution of the output given the distribution of the input (like the example above), for some examples you can not. Using a Monte Carlo simulation to approximate the output is indeed a way out of this.
If you are dealing with input that has only a small variance then you can use some polynomial approximation such as the Delta method. These approximations work well if the variance is not too large. See for instance an example here for the computation of a fraction $X/(X+Y)$ which can be solved with the Delta method, but has also an analytical solution.