Let $X\sim Poisson(m)$ and $m\sim\Gamma(2,1)$ then the join probability distribution function of $X$ and $m$ is given by $$\begin{align}f(x,m)&=f(x\mid x)\;f(m)\\&=\frac{m^xe^{-m}}{x!}\cdot\frac1{\Gamma(2)}me^{-m}\\&=\frac{m^{x+1}e^{-2m}}{x!}\end{align}$$
I understand how to find $f(m)$. Yet I have no idea how they got $f(x\mid m)$. Would someone be kind enough to show me the steps needed in order to arrive at that answer for $f(x\mid m)$?
The notation used here is obnoxious, using the same letter for different things. We have a function $m\mapsto f(m)$ and you're supposed to understand which function $f$ is by noticing that you're calling its argument $m$ rather than $x$. So if $m=3$, then you have $f(3)$. Does that mean "$f(m=3)$" or "$f(x=3)$"? When one writes $f_X(x)$, using capital $X$ as the random variable and $x$ as the argument to the function, then $f_X(3)$ is comprehensible and so is $\Pr(X\le x)$.
I'll use capital $M$ for the continuous random variable in this case and lower-case $m$ for the argument to its density function. Then $f_M(m) = me^{-m}$ for $m>0$, and $f_{X\,\mid\,M}(x) = \dfrac{m^x e^{-m}}{x!}$.
They ought to have said $X\mid M\sim\mathrm{Poisson}(M)$ rather than $X\sim\mathrm{Poisson}(M)$. The latter would mean the marginal distribution of $X$ is a Poisson distribution, and that is not true.
The thing that was written in obnoxious notation as $f(x\mid m)= \dfrac{m^xe^{-m}}{x!}$ should have been written as $f_{X\,\mid\,M}(x) = \dfrac{M^xe^{-M}}{x!}$ or as $f_{X\,\mid\, M=m}(x) = \dfrac{m^xe^{-m}}{x!}$.
As to where they got it: it was given that $X\mid M\sim\mathrm{Poisson}(M)$, and you know that the probability mass function for the Poisson distribution with expected value $M$ is $x\mapsto \dfrac{M^x e^{-M}}{x!}$.
Furthermore, the phrase "probability distribution function" is misleading, since that is often taken to mean the cumulative distribution function, and that is clearly not what is meant here.