Distributions of norm squared of multi-normal distribution had been extensively studied in radio-communication theory. See paper of S.O. Rice, "Distribution of Quadratic Forms in Normal Random Variables".
It is quite easy to find the characteristic function of $Z$, since
$$
\mathbb{E} \left( \mathrm{e}^{i t X_i^2} \right) = \frac{1}{\sqrt{1-2 i t}}
$$
Therefore
$$
\mathbb{E} \left( \mathrm{e}^{i t Z} \right) = \left( \prod_{i=1}^n 1-2 i t w_i \right)^{-1/2}
$$
Inversion in closed form is possible only in special cases.
We give two solutions. The first is a formal calculation. The second is informal, but more informative.
Formal verification: Let $X$ have exponential distribution with parameter $\lambda$. Let $k$ be a positive constant, and let $Y=kX$. The cumulative distribution function $F_Y(y)$ of $Y$, for $y \ge 0$, is given by
$$F_Y(y)=P(Y \le y)=P(X \le y/k)=F_X(y/k),$$
where $F_X$ is the cumulative distribution function of $X$.
Now there are several ways to find the density function $f_Y(y)$ of $Y$: (i) We could find $F_X(x)$ by integrating, substitute $y/k$ for $x$, and differentiate with respect to $y$; (ii) We could look up $F_X(x)$ and then proceed as in (i); (iii) Or else we can use the Fundamental Theorem of Calculus.
Let $x=y/k$. By the Chain Rule,
$$\frac{dF_X(y/k)}{dy}=\frac{dx}{dy}\frac{dF_X(x)}{dx}=\frac{1}{k}f_X(x)=\frac{1}{k}\lambda e^{-\lambda x}=\frac{\lambda}{k}e^{-\lambda y/k}=\lambda'e^{-\lambda' y},$$
where $\lambda'=\lambda/k$. That is what we wanted to show.
Note: The approach (iii) is in general the easiest, so that's what we used. It may look more difficult than approach (i), but that's because $\lambda e^{-\lambda x}$ is exceptionally easy to integrate. If we are working with the normal instead of the exponential, we can't even start on (i), while (iii) goes through just as easily as for the exponential.
Informal justification: The random variable $X$ might describe the lifetime, in days, of an atom of a certain radioactive substance $S$. Recall that the mean lifetime is $1/\lambda$.
On Planet $K$ far far away, the length of the day is one-third the length of an Earth day. So, on Planet $K$, and measuring time in $K$-days, substance $S$ will have mean lifetime $3/\lambda$ $K$-days. It follows that the lifetime $Y$ of an atom of substance $S$, as measured in $K$-days, has exponential distribution with parameter $\lambda/3$.
Similarly, if the length of the day on Planet $K$ is $1/k$ times an Earth day,
then on Planet $K$, measuring time in $K$-days, substance $S$ has mean lifetime $k/\lambda$. So the lifetime $Y$ of an atom of $S$, measured in $K$-days, has exponential distribution with parameter $\lambda/k$.
Best Answer
In other words, gives two independent random variables $X$ and $Y$, distributed according to hypoexponential distribution with parameters $\{ w_1, w_2, \ldots, w_n \}$ and $\{v_1,v_2, \ldots, v_m \}$ respectively, you are asking to determine the distribution of $Z=X-Y$.
Let $\Theta_X$, and $\Theta_Y$ denote matrices from the probability density functions of respective hypoexponential distributions, see wiki page: $$ f_X(x) = - \langle\vec{\alpha}_n, \exp(x \Theta_X) \Theta_X, \vec{1}_n \rangle \cdot [ x > 0 ] \qquad \qquad f_Y(y) = - \langle\vec{\alpha}_m, \exp(y \Theta_Y) \Theta_Y, \vec{1}_m \rangle \cdot [ y > 0 ] $$ where $(\alpha_n)_i = \delta_{i,1}$, $ (\vec{1}_n)_i = 1$ and $(\alpha_m)_j = \delta_{j,1}$, $ (\vec{1}_m)_j = 1$, $i=1,\ldots,n$, and $j=1,\ldots,m$.
Then $$ \begin{eqnarray} f_Z(z) &=& \int_{-\infty}^\infty f_X(z+y) f_Y(y) \mathrm{d} y = \int_{\max(-z,0)}^\infty f_X(z+y) f_Y(y) \mathrm{d} y \\ &=& \int_{\max(-z,0)}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(z+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{y \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\ &=& \int_{0}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(\max(z,0)+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{(\max(-z,0) + y) \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\ &=& \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{\max(z,0) \Theta_X} \otimes \mathrm{e}^{(\max(-z,0) ) \Theta_Y} \right) \cdot \left( \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y \right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \end{eqnarray} $$ The formula above tells the density function for $Z$ will be piecewise, much like Laplace distribution, with functional form of $X$ variate for $z>0$ and functional form of $Y$ variate for $z<0$.
Example: Consider an example with $n=2$ and $m=2$, and $\{w_1,w_2\} = \{1,2\}$, and $\{v_1,v_2\} = \{1,1\}$. Corresponding matrices are $$ \Theta_X = \left( \begin{array}{cc} -1 & 1 \\ 0 & -2 \end{array} \right) \qquad \Theta_Y = \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right) $$ Then $$ \exp\left(x \Theta_X \right) = \left( \begin{array}{cc} e^{-x} & e^{-x}-e^{-2 x} \\ 0 & e^{-2 x} \\ \end{array} \right) \qquad \exp\left(y \Theta_Y \right) = \left( \begin{array}{cc} e^{-y} & e^{-y} y \\ 0 & e^{-y} \\ \end{array} \right) $$ $$ \exp\left(x \Theta_X \right) \Theta_X = \left( \begin{array}{cc} -e^{-x} & 2 e^{-2 x}-e^{-x} \\ 0 & -2 e^{-2 x} \\ \end{array} \right) \qquad \exp\left(y \Theta_Y \right) \Theta_Y = \left( \begin{array}{cc} -e^{-y} & e^{-y}-e^{-y} y \\ 0 & -e^{-y} \\ \end{array} \right) $$ Using Kronecker product, $$ \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y = \left( \begin{array}{cccc} \frac{1}{2} & -\frac{1}{4} & -\frac{1}{6} & \frac{7}{36} \\ 0 & \frac{1}{2} & 0 & -\frac{1}{6} \\ 0 & 0 & \frac{2}{3} & -\frac{4}{9} \\ 0 & 0 & 0 & \frac{2}{3} \\ \end{array} \right) $$ Combining things, with little algebra we get: $$ f_Z(z) = \left\{ \begin{array}{cc} \frac{5}{18} & z=0 \\ \frac{1}{18} \mathrm{e}^{-z} \left(9-4 \mathrm{e}^{-z}\right) & z>0 \\ \frac{1}{18} \mathrm{e}^z (5-6 z) & z < 0 \\ \end{array} \right. $$