The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$.
The transformation theorem for probability densities states that:
$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$
where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.
(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector)
We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian
$$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$
which has determinant $\frac12$ for all $z,w$. We now plugin, and get
$$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$
for all $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute
$$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$
You wouldn't be able to find their joint pdf $f_{X,Y}$ given just their individual pdfs if they are not independent. You would need at least a conditional pdf or the joint pdf itself to know more about the relationship of their distributions. The joint pdf is related to the conditional pdf by
$$\begin{split}f_{X|Y}(x|y)&=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\\text{or} f_{Y|X}(y|x)&=\frac{f_{X,Y}(x,y)}{f_X(x)}\end{split}$$
If the variables are independent
$$\begin{split}\frac{f_{X,Y}(x,y)}{f_Y(y)}&=f_{X|Y}(x|y)\\
&=f_X(x)\end{split}$$
which is why you can directly multiply them together.
Best Answer
Just apply the fundamental theorem of calculus and the definition for probability density function::-
$\begin{align}\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^\infty h(a,y)\,\mathrm d y&= \int_{-\infty}^\infty\dfrac{\partial h(a,y)}{\partial a\qquad}\,\mathrm dy\\[2ex]\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^a g(x)\,\mathrm d x &= g(a)\\[2ex]&\text{so...}\\[3ex] f_{\small X+Y}(a) &=\dfrac{\mathrm d~~}{\mathrm d a}\mathsf P(X\leq a-Y)\\[1ex]&=\dfrac{\mathrm d~~}{\mathrm d a}\int_{-\infty}^\infty\int_{-\infty}^{a-y} f_{\small X, Y}(x,y)\,\mathrm d x\,\mathrm d y\\[1ex]&~~\vdots\end{align}$
$X+Y$ is a single random variable. It's probability density function would therefore be monovariate. The token $a$ was chose by the author to be the argument of this function.