Suppose we have $v$ and $u$, both are independent and exponentially distributed random variables with parameters $\mu$ and $\lambda$, respectively. 

How can we calculate the pdf of $v-u$?

I too prefer to call the random variables $X$ and $Y$.  You can think of $X$ and $Y$ as waiting times for two independent things (say $A$ and $B$ respectively) to happen.  Suppose we wait until the first of these happens.  If it is $A$, then (by the lack-of-memory property of the exponential distribution) the further waiting time until $B$ happens still has the same
exponential distribution as $Y$; if it is $B$, the further waiting time until $A$ happens still has the same exponential distribution as $X$.  That says that the conditional distribution of $X-Y$ given $X > Y$ is the distribution of $X$, and the conditional distribution of $X-Y$ given $X < Y$ is the distribution of $-Y$.  Since $P(X>Y) = \frac{\lambda}{\mu+\lambda}$, that says the PDF for $X-Y$ is
$$ f(x) = \frac{\lambda \mu}{\lambda+\mu}
\cases{e^{-\mu x} & if $x > 0$\cr
       e^{\lambda x} & if $x < 0$\cr}$$