Your question was actually answered eight decades ago by P. V. Sukhatme. I'll explain how to prove the problem assuming that you are familiar with the fundamentals of probability theory and the relationships between the exponential, gamma and $\chi^2$ distributions. First, if you reviewed any textbook that addresses order statistics, then you will observed that the joint probability density function of the order statistics $X_{(1)}, \ldots, X_{(n)}$ is defined to be
$$
f(x_{(1)}, \ldots, x_{(n)}) = n! \prod^n_{i = 1} f(x_{(i)}).
$$
By integrating the above joint probability density function with respect to $x_{(n)}, \ldots, x_{(r + 1)}$, respectively, one will obtain
$$
g(x_{(1)}, \ldots, x_{(r)}) = \frac{n!}{(n - r)!} \prod^r_{i = 1} f(x_{(i)}) \left[1 - F(x_{(r)})\right]^{n - r} = (n - r + 1)! \prod^r_{i = 1} f(x_{(i)}) \left[1 - F(x_{(r)})\right]^{n - r},
$$
where $F(\cdot)$ is the corresponding cumulative distribution function. In the case of the exponential distribution with scale parameter $\lambda$, the above density is given by
$$
g(x_{(1)}, \ldots, x_{(r)}) = \dfrac{(n - r + 1)!}{\lambda^r} \exp\left(\lambda^{-1} T\right),
$$
where $T = \sum^{r}_{i = 1} X_{(i)} + (n - r) X_{(r)}$. Second, define the following transformations:
$$
\begin{array}{l}
S_1 = n X_{(1)} \\
S_2 = (n - 1) \left[X_{(2)} - X_{(1)}\right] \\
\vdots \\
S_{r - 1} = [n - (r - 1) + 1] \left[X_{(r - 1)} - X_{(r - 2)}\right] \\
S_{r} = [n - r + 1] \left[X_{(r)} - X_{(r - 1)}\right] \\
\end{array}
$$
Clearly,
$$
\sum^r_{i = 1} S_i = \sum^{r - 1}_{i = 1} X_{(i)} + (n - r + 1) X_{(r)} = \sum^{r}_{i = 1} X_{(i)} + (n - r) X_{(r)} = T,
$$
and
$$
\begin{array}{l}
X_{(1)} = \frac{S_1}{n} \\
X_{(2)} = \frac{S_2}{n - 1} + \frac{S_1}{n} \\
\vdots \\
X_{(r - 1)} = \frac{S_{r - 1}}{n - r + 2} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n} \\
X_{(r)} = \frac{S_{r}}{n - r + 1} + \frac{S_{r - 1}}{n - r + 2} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n} \\
\end{array}
$$
Note that $S_1, \ldots, S_r$ are called spacings in statistical literature. Third, acquire the joint probability density function of the transformations $S_1, \ldots, S_r$, which is simply
$$
h(s_1, \ldots, s_r) = g\left(\frac{S_1}{n}, \frac{S_2}{n - 1} + \frac{S_1}{n}, \ldots, \frac{S_{r}}{n - r + 1} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n}\right) \mathbf{J}
$$
where $\mathbf{J}$ is the Jacobian of transformation. Clearly, since
$$
\mathbf{J} = \left|
\begin{matrix}
\frac{1}{n} & 0 & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n - 1} & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n - 1} & \frac{1}{n - 2} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n} & \frac{1}{n - 1} & \frac{1}{n - 2} & \cdots & \frac{1}{n - r + 1} \\
\end{matrix}
\right| = \frac{1}{(n - r + 1)!},
$$
then the above joint probability density function of the transformations $S_1, \ldots, S_r$ is reduced to
$$
h(s_1, \ldots, s_r) = \frac{1}{\lambda^r} e^{-\lambda^{-1} T} = \left(\dfrac{1}{\lambda} e^{-\lambda^{-1} s_1}\right) \cdots \left(\dfrac{1}{\lambda} e^{-\lambda^{-1} s_r}\right),
$$
i.e. $S_1, \ldots, S_r$ are independent and identically-distributed random variables that follow $\mathrm{Exp}(\lambda^{-1})$. Which means $T = \sum^r_{i = 1} S_i$ follows $\mathrm{Gamma}(r, \lambda)$. Hence, $\frac{2T}{\lambda} \sim \chi^2_{2r}$.
References: P. V. Sukhatme (1937), Tests of significance for samples of the $\chi^2$ population with two degrees of freedom, Ann. Eugenics 8, 52-56.
