I would like to add another answer to this old question. Consider
the case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$:

$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$

Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in fact
we can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the *universal* point $S = R[T_0]$, $a_0 = T_0$, we see that the ``universal Taylor expansion'' of $f$ is
$$
f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T].
$$
If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as
$$
1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i
$$
Looking mod $(1\otimes T-T\otimes 1)^2$ we get:
$$
1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2}
$$

Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by
$1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear terms
of Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function
$f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$.

This also explains nicely what happens in higher degree. We can introduce the algebras
$P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the *ring of Taylor expansions of degree $\leq n$* where
the terms of degree at most $n$ of the Taylor expansion live. There is a natural map
$d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion.

This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), because
locally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and this
map induces an isomorphism on $I/I^2$ and $P^n$ more generally.