A little late to the party, but here's what I think happens.
I will assume $X$ is closed and oriented, with fundamental class $[X]\in H_n(X)$. I will also assume we're using field coefficients (or $X$ has torsion-free homology) so there is an isomorphism $H_*(X)\otimes H_*(X)\cong H_*(X\times X)$ given by cross product.
Let $F\colon X\to X\times Y$ be the map $F(x)=(x,f(x))$. Then $\Delta_f=F_*[X]\in H_n(X\times Y)$. Note that there is a factorisation $F=(1\times f)\circ d$, where $d\colon X\to X\times X$ is the diagonal map.
Under our assumptions, a basis $\{b_i\}_{i\in I}$ for $H_*(X)$ has a Poincaré dual basis $\{b_i'\}$, and the diagonal class is given by $d_*[X]=\sum_{i\in I} b_i\times b_i'$ (see Chapter 13 of Milnor and Stasheff). Hence
\[ \Delta_f = (1\times f)_*d_*[X] = \sum_{i\in I} b_i \times f_*(b_i'). \]
So it's not enough to know just the homology class represented by $f$, you have to know the whole induced homology homomorphism.
With regards your second question, as noted in the comments, $f$ is a covering map. The map $F\colon X\to X\times Y$ is an embedding, whose normal bundle is isomorphic to the tangent bundle of $X$. So this self-intersection number $\Delta_f^2$ should just be the Euler characteristic of $X$.