Fundamental group of two tori with a circle ($S^1✕${$x_0$}) identified
Here are two tori with a copy of $S^1$ identified. The picture is clearer when we move $x_0$ to the inner rim of one torus, and to the outer rim on the other torus.
The two images below show a path on one of the tori that belongs to the homotopy class of the latter generator.
Basically Van Kampen says that these two become homotopic in $\pi_1$ of the combination. Geometrically this is obvious, as you can slide that path from one torus to the other via the shared circle of contact.
More precisely, if $b$ denotes the homotopy class above and $a_1$ (resp. $a_2$) is a loop around the tube of the bigger (resp. smaller) torus, then elements of the amalgamated product are objects like $$ (a_1^{k_1},b^{\ell_1})*(a_2^{m_1},b^{n_1})*\cdots*(a_1^{k_t},b^{\ell_t})*(a_2^{m_t},b^{n_t}), $$ where the exponents $k_i,\ell_i,m_i,n_i$ are all arbitrary integers, $t$ is a natural number and factors with $a_1$ (resp. $a_2$) alternate.
Because $b$ commutes with both $a_1$ and $a_2$, the above product simplifies to $$ a_1^{k_1}a_2^{m_1}a_1^{k_2}a_2^{m_2}\cdots a_1^{k_t}a_2^{m_t}b^r $$ with $$ r=\sum_{i=1}^t(\ell_i+n_i). $$ This is path that first does $k_1$ laps around the outer tube, then $m_1$ laps around the inner loop et cetera, and at some point also goes $r$ laps along the intersection.
Looks like $(\Bbb{Z}*\Bbb{Z})\oplus\Bbb{Z}$ as a group. A direct product of two groups where the first factor is a free product of two infinite cyclic groups, and the latter is just $\Bbb{Z}$.
In retrospect this is clear. Undoubtedly you know that the fundamental group of figure 8 is that free product of two infinite cyclic groups. Your space is the direct product $8\times S^1$.
I'll add that this means the group is isomorphic to $\mathbb{Z} \times F$, where $F$ is a free (nonabelian) group on two generators.
In particular, before modding out, there are two pairs of generators, say $\{x,y\}$ for the first $\mathbb{Z}\times \mathbb{Z}$ and $\{a,b\}$ for the other. Each pair commutes with itself, but the two pairs do not commute. Then, modding out by $N$ just says $y = b$.
So now we have three generators, $y(=b), x, a$. The first one commutes with both the other two, and $x$ and $a$ don't commute. That gives $\mathbb{Z} \times F$.