homology functor from abelian groups to abelian groups given a topological space $X$, $H_n(X,-)$ is a functor from the category of abelian groups to itself. i want to clarify the following :
1) given an  homomorphism  $f:G\rightarrow H$  of abelian groups what is explicitly the induced map $f_*:H_n(X,G)\rightarrow H_n(X,H)$
my guess: an element in $H_n(X,G)$ is a formal sum $\sum{g_i c_i}$ where $g_i\in G$ and $c_i$ a class of a cyle in $C_n(X,G)$ the free $\mathbb Z$-module on $n$-singular simplices.
so $f_*(\sum{g_i c_i})=\sum{f(g_i) c_i}$
2)given an abelian group $A$, what is the canonical homomorphism $f:\mathbb Z \rightarrow A$ that is used to induce  $f_*:H_n(X,\mathbb Z)\rightarrow H_n(X,A)$ and then induce $$f_{**}:H_n(X,\mathbb Z)\otimes A\rightarrow H_n(X,A)$$ that gives the short exact sequence in the universal coefficient theorem in homology:
$$0 \rightarrow H_i(X, \mathbb{Z})\otimes A\rightarrow H_i(X,A)\rightarrow\mbox{Tor}(H_{i-1}(X, \mathbb{Z}),A)\rightarrow 0$$
 A: 1) Yes, your guess is correct.
2) There is no non-zero canonical morphism $\mathbb Z\to A$.
The map $\phi:H_\bullet(X,\mathbb Z)\otimes A\to H_\bullet(X,A)$ is constructed differently. Let $\alpha\otimes a$ be an elementary tensor in the domain. The homology class $\alpha$ is the class of some cycle $\sum_ic_i\sigma_i$ with $c_i\in\mathbb Z$. Then $\phi(\alpha\otimes a)$ is the class in $H_\bullet(X,A)$ of the element $\sum_ic_ia\sigma_i$. You can easily check that this is in fact a cycle in the complex which computes $H_\bullet(X,A)$, so this makes sense.
 
The very best to understand what are the maps in the Universal Coefficient Theorem is to follow a proof of the theorem in detail: all proofs I know of actually construct the maps!
 
Later: I claimed there is no non-zero canonical map $\mathbb Z\to A$. Let me prove at least there is no natural map (for there is no sensible definition of canonical!). Suppose for all abelian groups $\phi_A:\mathbb Z\to A$ is a group homomomorphism which depends naturally on $A$. Let $A$ be a group, and let $i_1,i_2:A\to A\oplus A$ be the two obvious inclusions into the "coordinate axes". Then naturality implies that $$i_1(\phi_A(1))=\phi_{A\oplus A}(1)=i_2(\phi_A(1)),$$ so $\phi_{A\oplus A}(1)\in i_1(A)\cap i_2(A)=0$. It follows that $\phi_{A\oplus A}(1)=0$ and, since $i_1(\phi_A(1))=\phi_{A\oplus A}(1)$ and $i_1$ is injective, that $\phi_A(1)=0$. Thus $\phi_A=0$.
A: The first point is essentially correct.  Let $C_n(X)$ be the $n$-chains in $X$ with integral coefficients.  For an abelian group $A$, by definition, $C_n(X,A) = C_n(X) \otimes A$.  To the chain complex $(C_*(X), \partial)$ there is an associated chain complex $(C_*(X,A), \partial \otimes 1_A)$, where $1_A$ is the identity map on $A$.  The homology of the first complex is, by definition, $H_*(X)$ (with integral coefficients), and the homology of the second complex is, by defintion, $H_*(X,A)$.  
Given a homomorphism $f:A \to B$ of abelian groups, there is an induced map 
$f_*: H_*(X,A) \to H_*(X,B)$ given by $f_*(\sum \sigma_i \otimes a_i) = \sum \sigma_i \otimes f(a_i)$.  Here the notation is a little sloppy:  I am regarding $\tau = \sum \sigma_i \otimes a_i$ as an element of $C_n(X,A)$, not as a homology class.  So, you need to check that this map is well-defined.  (This is a highly recommended exercise.)
As to the second point, the universal coefficient theorem is comparing $H_n(X) \otimes A$ with $H_n(X,A)$.  The difference is the order in which we tensored with $A$.  The map $H_n(X) \otimes A \to H_n(X,A)$ is... 
(EDIT: thanks to Mariano's remarks below) given by mapping $[\tau] \otimes a$ to the class $[\tau \otimes a]$, where $\tau$ is a cycle in $C_n(X)$ and $[\tau]$ is the homology class that it represents.
(Here is my original, incorrect statement:
"induced by the map on the chain level:  $C_n(X) \otimes A \to C_n(X) \otimes A$, i.e. the identity map.")
Again, there is work to be done: you need to check that this induces a well-defined map on homology.
To get a feel for what is going on, it makes sense to have an example at hand.  Comparing $H_2(\mathbb{R}P^2)$ to $H_2(\mathbb{R}P^2, \mathbb{Z}_2)$ should be helpful.
