Intuition for Haar measure of random matrix
You want to think of the Haar measure $d\mu(U)$ as a way of measuring uniformity in the group $U(N)$ of unitary $N\times N$ matrices.
To form your intuition, consider $N=1$. You then have $U=e^{i\phi}$, with $0<\phi\leq 2\pi$ and $d\mu(U)=d\phi$ measures the perimeter of the unit circle. This is a uniform measure, because $d(\phi+\phi_0)=d\phi$ for any fixed phase shift $\phi_0$. You could write the requirement of uniformity in the form $d\mu(UU_0)=d\mu(U)$, with $U_0=e^{i\phi_0}$ the unitary matrix corresponding to the phase shift $\phi_0$.
Once your intuition is formed for $N=1$, you simply generalize to $N>1$ using the same definition of uniformity, $d\mu(UU_0)=d\mu(U)$ for any fixed $U_0\in U(N)$. For orthogonal (or symplectic) matrices you use the same definition of uniformity, with $U_0$ now restricted to the orthogonal or symplectic subgroup of $U(N).$
To explicitly write down the Haar measure $d\mu(U)$ in terms of the matrix elements of $U$ is only easily done for a few small values of $N$. (In particular, there is no relationship to random directions of rows or columns, as Yemon Choi pointed out.) You typically do not need such explicit expressions, since integrals with the Haar measure can be evaluated by using only the definition of uniformity.
In response to the follow-up question: If you wish to evaluate Haar-measure integrals of polynomials of matrix elements of $U$, you can use the socalled Weingarten functions.
http://en.wikipedia.org/wiki/Weingarten_function
Here is a Mathematica program to generate these,
http://arxiv.org/abs/1109.4244
If you need an explicit expression for the Haar measure, the steps to take are the following:
1) parameterize your matrix $U$ in terms of a set of real parameters $\{x_i\}$.
2) calculate the metric tensor $m_{ij}$, defined by $\sum_{ij}|dU_{ij}|^2 = \sum_{ij}m_{ij}dx_i dx_j$
3) obtain the Haar measure by equating $d\mu(U) = ($Det $m)^{1/2}\prod_i dx_i$
This is the general recipe. In practice, for many parameterizations the answer is in the literature. In particular, for the Haar measure in Euler angle parameterizations see:
http://arxiv.org/abs/math-ph/0205016
http://www.cft.edu.pl/~karol/pdf/ZK94.pdf
Dear Jiahao Chen, if you're interested in explicit expressions of the Haar measure you might want to look at the paper http://arxiv.org/abs/1103.3408 which contains simple parametrizations of U(N) and SU(N), as well as a formula of the normalized Haar measure for arbitrary N. This paper could be interesting for you as the provided framework can directly be applied to compute group integrals. Knowledge of the Weingarten functions is not required.
I like the following characterisation. Consider a continuous function $f:G\to\mathbb{R}$ (for any compact Lie group $G$) and the set $T=\{t_g(f): g\in G\}$ of translates of $f$, where $t_g(f)(x)=f(gx)$. Let $C$ be the closure of the convex hull of $T$. It can be shown that $C$ contains a unique constant function, and the value of that function is the inegral of $f$ with respect to Haar measure.