What is the "natural homomorphism" in the definition of an *essential manifold*? The following definition of "essential manifold" is in this wiki page:

A closed $n$-manifold $M$ is called essential if its fundamental class $[M]$ defines a nonzero element in the homology of its fundamental group $\pi$, or more precisely in the homology of the corresponding Eilenberg–MacLane space $K(\pi, 1)$, via the natural homomorphism
$$\displaystyle H_{n}(M) \to H_{n}(K(\pi ,1)).$$
Here the fundamental class is taken in homology with integer coefficients if the manifold is orientable, and in coefficients modulo $2$, otherwise.

My question is: what is this "natural homomorphism" in the definition?
Added:
In the original paper of Gromov (Gromov, M.: "Filling Riemannian manifolds," J. Diff. Geom. 18 (1983), 1–147.), he defines an essential manifold apparently in a different way:

He takes $K$ to be an arbitrary $K(\pi, 1)$ or $\pi = \pi_1(M,m)$?
Question: are both definitions equivalent?
 A: Assuming that $M$ is connected, there is a unique, up to homotopy, map $f: M\to X=K(\pi,1)$ inducing the isomorphism $\pi_1(M, m)\to \pi_1(X, x)=\pi$. Then $f$ induces a homomorphism of homology groups
$$
H(f): H_*(M)\to H_*(X).
$$ 
This is the natural homomorphism appearing in Gromov's definition. 
A: I will address your second question, to show that these two definitions are equivalent. One direction is automatic since $K(\pi, 1)$ is an aspherical space. 
Suppose $X$ and $K$ are connected CW complexes and suppose $K$ is aspherical. All spaces are assumed to be pointed and by "map" I will mean "pointed continuous function". 
For any map $f\colon X \to K$ I claim it admits a factorization (up to homotopy) through $K(\pi_1(X), 1)$. Note that $K(\pi_1(X), 1)$ can be constructed so that there is a canonical isomorphism between its fundamental group and $\pi_1(X)$, so I will be implicitly using this identification even though strictly speaking they are not equal.

Black Box: if $X$ and $K$ are connected CW complexes and $K$ is apsherical, then for any group homomorphism $\varphi \colon \pi_1(X) \to \pi_1(K)$ there is a homotopically unique map $f\colon X \to K$ such that $\pi_1(f) = \varphi$. 

(This result is typically proven using obstruction theory and I will omit a proof because it is standard.)
By the black box there is a homotopically unique map $i\colon X\to K(\pi_1(X), 1)$ inducing "the identity" on $\pi_1$, and a homotopically unique map $h\colon K(\pi_1(X),1) \to K$ which induces "$\pi_1(f)$". Then the composition
$$ h\circ i \colon X \to K(\pi_1(X), 1) \to K $$
induces the same homomorphism on $\pi_1$ as $f$, therefore by uniqueness $f \sim h \circ i$.
Now to deduce the result from the claim, if there is an $n\in \mathbb{N}$ and $a \in H_n(X)$ such that $H_n(f)(a) \neq 0\in H_n(K)$, then by homotopy-invariance of homology $H_n(h\circ i)(a) \neq 0$ so $H_n(i)(a)$ must be non-zero in $H_n(K(\pi_1(X), 1))$ as well, so we have the other direction of the equivalence.
