In Riemann's paper he defined $\xi(t)=\Pi(\frac{s}{2})(s-1)\pi^{-\frac{s}{2}}\zeta(s)$, where $s=\frac{1}{2}+ti$. On page 4 he said:
The number of roots of $\xi(t)=0$, whose real parts lie between $0$ and $T$, is approximately $=\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}$, because the integral $\int d\log\xi(t)$, taken in a positive sense around the region consisting of the values of $t$ whose imaginary parts lie between $\frac{1}{2}i$ and $-\frac{1}{2}i$ and whose real parts lie between $0$ and $T$, is (up to a fraction of the order of magnitude of the quantity $\frac{1}{T}$) equal to $(T\log\frac{T}{2\pi}-T)i$.
How did he compute $\int d\log\xi(t)$?
Because the integral
$$ \int\mathrm d\log\xi(s)\tag1 $$
is taken around a rectangular region symmetric about the critical line $\Re(s)=\frac12$, we can use the functional equation $\xi(s)=\xi(1-s)$ (the modern definition) to simplify the integral so that computing (1) only requires the calculation of
$$ \left(\int_{\frac12}^2+\int_2^{2+iT}+\int_{2+iT}^{\frac12+iT}\right)\mathrm d\log\xi(s).\tag2 $$
Because analytic functions have isolated zeros, we can choose $T$ such that no zero satisfies $\Im(s)=T$.
To compute this integral, we unambiguously can define $\log\xi(\sigma+it)$ by continuous variation from $s=2$ and $s=2+it$ to $s=\sigma+it$ for $\frac12\le\sigma\le2$ and $0\le t\le T$ so that
$$ \log\xi(s)=\log\left[\frac12s(s-1)\right]-\frac s2\log\pi+\log\Gamma\left(\frac s2\right)+\log\zeta(s)\tag3 $$
is valid on the path of integration of (2). Plugging (3) into (2) and apply Stirling's approximation to the $\Gamma$-function, we obtain
$$ N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+O(1)+R(T), $$
where $R(T)$ is, up to a constant multiple, the remaining integral over $\mathrm d\log\zeta(s)$. Currently, it is known that $R(T)$ is of $O(\log T)$, so combining everything gives
$$ N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+O(\log T). $$
For a detailed proof of this asymptotic formula, see my two-year old blog.