I recall enjoying the van Kampen exercises in grad school, so I will give this one a try. The topologists can hopefully add more helpful answers.
As the first example consider the sphere with a single handle, i.e. the torus. You hopefully already know the answer, so let's see how van Kampen does it. We are to split the torus into two parts, $U$ and $V$ with a path-connected intersection, such that we know the fundamental group of $U,V$ and $U\cap V$. Let $V$ be just a small open disk on the surface of the torus. Let $K$ be an even smaller closed disk inside $V$, and let $U$ be the complement of $K$ on the entire torus, so $U\cap V = V\setminus K$ is an open annulus around the perimeter of $V$.
$V$ is contractible and has a trivial fundamental group.
$U\cap V$ contracts to a circle, and has the infinite cyclic group as the fundamental group. A generator of this group is the loop $g$ going once around the circle.
$U$ is essentially the torus with a hole in it made by the removal of the patch $K$. By making that hole bigger and bigger, we eventually see that $U$ contracts to a figure eight 8. If we look at the torus as a bicycle tube, then $U$ is the tube with the valve and its surroundings removed, and it contracts to the union of a big circle $x$ (the points that would touch the ground, if you rotate the wheel 360 degrees) and one small circle $y$ around the tube. These two circles intersect at a single point. So the fundamental group of $U$ is the free group on two generators $x$ and $y$.
What does van Kampen tell us about the fundamental group of the union $U\cup V$? Basically we get the free product of the fundamental groups of $U$ and $V$, but we need to do a bunch of identifications by introducing relations that equate the images (under the induced by the inclusion map) of the elements of $\pi_1(U\cap V)$ on either side.
Here $\pi_1(U\cap V)=\langle g\rangle$, so we only need to check, what kind of a relation we get by equating the image of $g$ in $\pi_1(U)$ and $\pi_1(V)$. In $\pi_1(V)$ the image of $g$ is, of course, trivial, because the loop trivially contracts to a point, once we allow it to pass into $K$. The fun part is to observe that when we "expand the hole $K$ to the complement of the figure 8", the loop $g$ becomes the commutator $xyx^{-1}y^{-1}$ (alter the order of the factors depending on the choices of the orientations that you made).
You may need to draw a picture to see this happening. As a substitute I would suggest that if we form the torus by glueing together the opposite sides of a rectangle in the usual way, then $K$ is a hole in the middle, $g$ is a loop around it, the figure 8 is the border of the rectangle with the obvious identifications, and "expanding the hole" results in pressing the loop to go once around the perimeter of the square.
Anyway, van Kampen tells us now to identify $xyx^{-1}y^{-1}$ with the trivial element turning the free group on two generators into a free abelian group on two generators. The powers of $g$ don't introduce any new relations, so we are done.
Best Answer
I just consider the case $h=3$, but the argument is completely general.
From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure:
Now let $U, V \subset N_3$ be subspaces illustrated by the figure below:
Applying van Kampen theorem, we deduce that $\pi_1(N_3)$ is the quotient of the free product $\pi_1(U) \ast \pi_1(V)$ identifying the images of $\pi_1(U \cap V)$ into $\pi_1(U)$ and $\pi_1(V)$. In fact, $\pi_1(V)$ is clearly trivial so that $\pi_1(N_3)$ turns out to be the quotient of $\pi_1(U)$ by its subgroup corresponding to $\pi_1(U \cap V)$.
Now, there is naturally a deformation retract $\phi$ from $U$ to the quotient of $\partial P$: it is just a graph $\Gamma$, more precisely a bouquet of three circles labelled by $a_1$, $a_2$ and $a_3$. Therefore, $\pi_1(U)$ is just the free group $\langle a_1,a_2,a_3 \mid \ \rangle$.
Then, there is a deformation retract from $U \cap V$ to the circle $\partial V$. Therefore, $\pi_1(U \cap V)$ is infinite cyclic and its image into $\pi_1(U)$ is the homotopy class of the circle $\phi(\partial V)$ in the quotient $\Gamma$ of $\partial P$. Such a loop in $\Gamma$ is labelled by $a_1^2a_2^2a_3^2$.
Finally, we conclude that $$\pi_1(N_3)= \langle a_1,a_2,a_3 \mid a_1^2a_2^2a_3^2 \rangle.$$