Y'know, for your portion of the audience, I have a far wider range of expression:

We all know and love the topological homotopy, most naïvely expressed as a continuous function H(x,t):X x I -> Y such that H(x,0)=f(x) and H(x,1)=g(x).

We also all know and love the chain homotopy h, given as a degree -|d| map X -> Y such that f-g = dh+hd.

Now, the funky thing appears when we recognize that the chain complex most naturally associated with the topological construction is, in fact, I=0 -> kc -> ka(+)kb -> 0 with the single non-trivial differential c -> a-b.
So we'll consider X (x) I as a chain complex. This has in a given degree n the composition
X_(n-1) (x) I_1 (+) X_n (x) I_0
Now, the statement that would be neat to have proven (and that got proven in the seminar in question) is that existence of a chain homotopy h: f ~ g is equivalent to existence of a chain map H: X (x) I -> Y such that H(x,a) = f(x), H(x,b) = g(x).

Firstly, lets suppose we have H and want to find a h. Well, consider the function (curry if you wish) h = H(-,c): X -> Y. We'll want to take a look at what happens if we take a close look at dh+hd.
So
dh(x) = dH(x,c)=Hd( x(x)c )=H(dx(x)c - x(x)dc) = H(dx(x)c) - H( x(x)a - x(x)b ) =
= h(dx)-(H(x,a)-H(x,b)) = hd(x) -(f(x)-g(x))
and thus (up to me screwing up signs), dh+hd=f-g.
Suppose now instead we have a h and seek a H. Well, define H by
H(x,a) = f(x)
H(x,b) = g(x)
H(x,c) = h(x)
Now
dH(x,a)=df(x)=fd(x)=H(dx,a)-H(x,da)=Hd(x,a) (where da=0 and db=0 by the differential in I)
dH(x,b)=Hd(x,b) by the same argument and
dH(x,c)=dh(x)=f(x)-g(x)-hd(x)=H(x,a)-H(x,b)-H(dx,c)=H(x,dc)-H(dx,c)=Hd(x,c) (again with fudging the signs that don't bother coming out just right)

So, this juggling shows us that the chain homotopy and topological homotopy, really, really are the same thingie.