Theorem 70.1 (Seifert-van Kampen theorem). Let $X=U \cup V$, where $U$ and $V$ are open in $X$; assume $U, V$, and $U \cap V$ are path connected; let $x_0 \in U \cap V$. Let $H$ be a group, and let $ \phi_1: \pi_1\left(U, x_0\right) \longrightarrow H \ \text { and } \ \phi_2: \pi_1\left(V, x_0\right) \longrightarrow H $ be homomorphisms. Let $i_1, i_2, j_1, j_2$ be the homomorphisms indicated in the following diagram, each induced by inclusion.
If $\phi_1 \circ i_1=\phi_2 \circ i_2$, then there is a unique homomorphism $\Phi: \pi_1\left(X, x_0\right) \rightarrow H$ such that $\Phi \circ j_1=\phi_1$ and $\Phi \circ j_2=\phi_2$.
This theorem says that if $\phi_1$ and $\phi_2$ are arbitrary homomorphisms that are "compatible on $U \cap V$," then they induce a homomorphism of $\pi_1\left(X, x_0\right)$ into $H$.
Proof. Uniqueness is easy. Theorem 59.1 tells us that $\pi_1\left(X, x_0\right)$ is generated by the images of $j_1$ and $j_2$. The value of $\Phi$ on the generator $j_1\left(g_1\right)$ must equal $\phi_1\left(g_1\right)$, and its value on $j_2\left(g_2\right)$ must equal $\phi_2\left(g_2\right)$. Hence $\Phi$ is completely determined by $\phi_1$ and $\phi_2$. To show $\Phi$ exists is another matter!
Which by generating in theorem 59.1 it means :
given any loop $f$ in $X$ based at $x_0$, it is path homotopic to a product of the form $(g_1 * (g_2 * (\ldots * g_n)))$, where each $g_i$ is a loop in $X$ based at $x_0$ that lies either in $U$ or in $V$.
I don't see exatly why the value of $\Phi$ on the $j_1(g_1)$ must equal $\phi_1(g_1)$?
This is the obvious consequence of the property that $\Phi \circ j_k = \phi_k$ for $k = 1, 2$.
$\pi_1(X,x_0)$ is generated by the set $\Gamma = j_1(\pi_1(U,x_0)) \cup j_2(\pi_1(V,x_0))$. For each $g_1 \in \pi_1(U,x_0)$ we have $\Phi(j_1(g_1)) = (\Phi \circ j_1)(g_1) = \phi_1(g_1)$ and for each $g_2 \in \pi_1(V,x_0)$ we have $\Phi(j_2(g_2)) = (\Phi \circ j_2)(g_2) = \phi_2(g_2)$.
Hence $\Phi \mid_\Gamma$ is uniquely determined by $\phi_1$ ad $\phi_2$.