I think the origin of the confusion in the comments is that Munkres denotes loops as maps from intervals, instead of maps from $S^1$. A map $\sigma : [0, 1] \to X$ with $\sigma(0) = \sigma(1) = x_0$ is, however, equivalent to a map $\gamma : (S^1, b_0) \to (X, x_0)$ coming from the universal property of quotient spaces, where the domain circle is the quotient space $[0, 1]/\!\!\sim$, $\sim$ pasting $0$ and $1$ to the point $b_0$. Nevertheless, I'll use the notation you're comfortable with :
$\gamma : [0, 1] \to S^1$ be any loop in $S^1$ based at $\gamma(0) = \gamma(1) = b_0$. As $h_*$ is trivial, $h \circ \gamma : [0, 1] \to X$ based at $h(\gamma(0)) = h(\gamma(1)) = x_0$ can be path-homotoped to the zero loop $e_{x_0}$ at $x_0$. Let $F : [0, 1]\times [0, 1] \to X$ denote the path homotopy between $h \circ \gamma$ and $e_{x_0}$.
The desired nullhomotopy then is $H : S^1 \times [0, 1] \to X$ obtained from the universal property of quotient maps, where $S^1 \times [0, 1]$ is considered as the quotient space $[0, 1] \times [0, 1]/\!\! \sim$ under the identification $(0, x) \sim (1, x)$ for all $x$.
Geometrically, this means we are taking the loop in $(X, x_0)$, and letting the intermediates of the nullhomotopy to be the intermediates of the path-homotopy between $\gamma$ and the zero loop. That's all there is to it, as Qiaochu mentioned in the comments.
Indeed, Theorem 1B.9 is a good tool here. In particular, you'll want to use the uniqueness part of the theorem. That means that any two maps $(X,x_0)\to (K(G,1),y_0)$ which induce the same map on $\pi_1$ are homotopic. What does that tell you about a map $f:(X,x_0)\to (K(G,1),y_0)$ which induces the trivial homomorphism on $\pi_1$?
The answer is hidden below.
It tells you $f$ is homotopic to the constant map $(X,x_0)\to (K(G,1),y_0)$, since the constant map also induces the trivial homomorphism on $\pi_1$. So, if there are no nontrivial homomorphisms $\pi_1(X)\to G$, this applies to every map $f:X\to Y$ (for an appropriate choice of basepoints $x_0$ and $y_0$).
Best Answer
$\pi_1(X)$ consists of homotopy classes of based maps $S^1 \to X$. The group $\pi_1(S^1) \cong \mathbb{Z}$, where the class of identity map $id: S^1 \to S^1$ is the generator of $\pi_1(S^1)$.
Then $h_* : \pi_1(S^1) \to \pi_1(X)$ sends the generator $[id]$ of $\pi_1(S^1)$ to $[h \circ id] = [h] \in \pi_1(X)$. The triviality of $h_*$ is then equivalent to $[h] = [*] \in \pi_1(X)$, where $* : X \to X$ is the constant map at the basepoint. This is precisely the notion that $h$ is (based) nullhomotopic.