For the above question, I can solve it quite easily by utilising a counting process however I'm curious as to how one would go solving it by utilising the exponential distribution.
Namely, I'm stuck at understanding how to calculate the probability that the amount of time elapsed since the 2nd customer arrives is greater than 10 minutes given that one customer arrives within 10 minutes i.e. Pr(V1 + V2 > 10 | V1 <= 10)
How do we figure out this probability?
If $V_1,V_2,\ldots$ are i.i.d. $\mathsf{Expo}(\lambda)$ random variables, it is straightforward to show by induction that $S_n := \sum_{i=1}^n V_n$ has Erlang distribution with density $$ f_n(t) = \frac{\lambda(\lambda t)^{n-1}}{(n-1)!} e^{-\lambda t},\ t>0. $$ It follows then that $$ \mathbb P(S_n \leqslant T) = \int_0^T \frac{\lambda(\lambda t)^{n-1}}{(n-1)!} e^{-\lambda t}\ \mathsf dt = 1-\sum_{k=0}^{n-1}\frac{(\lambda T)^n}{k!} e^{-\lambda T}.\tag 1$$ Here $n=2$, $\lambda = \frac1{10}$, and $T=10$, and so the desired probability is $1-2e^{-1}$.
Alternatively (as Henry suggested), you could compute this probability by a double integral of the joint density of $(V_1,V_2)$:
\begin{align} \mathbb P(V_1+V_2\leqslant T) &= \iint_{\{(s,t)\in (0,T)^2 : s<t\}} f_{V_1,V_2}(s,t)\ \mathsf ds\ \mathsf dt\\ &= \int_0^T \int_0^{T-t} \lambda e^{-\lambda s}\lambda e^{-\lambda t}\mathsf ds\ \mathsf dt\\ &= 1-(1+\lambda T)e^{-\lambda T}, \end{align} which is the same as $(1)$ when $n=2$.