You *don't* need to assume the Leibniz rule, but can use a linearity property instead. I addressed this in comments to my answer at Different ways of thinking about the derivative, but this may be a good place to paste those comments together into a single answer.

The set of all derivations at a point $P$ on a manifold can be thought of as the dual space of $M/M^2$, where $M$ is the maximal ideal of smooth functions vanishing at $P$. That description of the tangent space at $P$, as a dual space to $M/M^2$, makes no mention at all about the product rule, but from the *special features* of the manifold setting the linearity that is built into the definition of a dual space can indeed be converted into a derivation (something linear with a product rule).

Here's a general setup. Let $A$ be a local ring with maximal ideal $M$ and set $F = A/M$. (Think $A$ = local ring of smooth functions at point $P$ on a manifold and $M$ = the smooth functions on the manifold that vanish at $P$, so $F$ is essentially the field of real numbers as constant terms of series expansions for functions at $P$.) Let's now make an *assumption* that is true for local rings arising in the setting of manifolds but is not true for all local rings: the field $F = A/M$ naturally lies inside the ring $A$. More precisely, let's suppose there is a homomorphism $F \rightarrow A$ such that the composite $F \rightarrow A \rightarrow A/M$ is the identity. (For rings of functions at a point on a manifold this is very natural since the real numbers sit inside $A$ as constant functions. But not every local ring naturally contains its residue field, such as a local ring of char. 0 whose residue field has characteristic $p$.) We will show with this assumption that any $F$-linear map $f \colon M/M^2 \rightarrow F$ can be turned into a derivation $A \rightarrow F$.

From the assumption we get a direct sum decomposition $A = F \oplus M$. This direct sum decomposition is obvious geometrically for a smooth function $h(x)$ around a point $a$, since it comes from writing $h(x) = h(a) + (h(x) - h(a))$, with $h(a)$ a real number an $h(x) - h(a)$ vanishing at $a$. OK, now let $f$ be any element of the dual space of $M/M^2$, i.e., $f \colon M/M^2 \rightarrow F$ is an $F$-linear map (since $F = A/M$, $M/M^2$ is naturally an $F$-vector space). Saying the function $f$ is $F$-linear is the same as saying it's $A$-linear since both $M/M^2$ and $F = A/M$ are $A$-modules and in both cases $M$ multiplies on them like 0. Since $M$ is an $A$-module (it's an ideal in $A$!), an $A$-linear map $f \colon M/M^2 \rightarrow F$ can be pulled back to an $A$-linear map $M \rightarrow F$ which is 0 on $M^2$. Let us extend $f$ to a map $A \rightarrow F$ by just declaring it to act on the direct sum $A = F \oplus M$ as 0 on the $F$-summand. That is, we simply define
$$f(c+m) = f(m)$$ for any $c$ in $F$ (thought of as a subfield of $A$) and $m$ in $M$.
We are calling this extended function $f$ as well, so we have turned $f \colon M/M^2 \rightarrow F$ into a function $A \rightarrow F$ still denoted $f$.

Now we have a function $f \colon A \rightarrow F$ that is $A$-linear and it is 0 on both $F$ and $M^2$. It's time to see that $f$ satisfies the product rule! For $a$ and $b$ in $A$, write $a = c + m$ and $b = d + n$ where $c, d$ are in $F$ and $m, n$ are in $M$. Then $f(a) = f(m)$ and $f(b) = f(n)$. We have $$f(ab) = f((c+m)(d+n)) = f(cd+cn+dm + mn) = f(cd) + f(cn) + f(dm) + f(mn).$$ Since $f$ is 0 on $F$ and $M^2$, we get $f(ab) = f(cn) + f(dm) = cf(n) + df(m) = cf(b) + df(a)$.

The $f$-values are in $F = A/M$, so in $cf(b) + df(a)$ things only matter mod $M$. Since $a \equiv c \bmod M$ and $b \equiv d \bmod M$, $cf(b) + df(a) = af(b) + bf(a)$. Voila: $f(ab) = af(b) + bf(a)$, which is the product rule.

Summary: Let $A$ be a local ring with maximal ideal $M$ and assume there is an *embedding* $A/M \rightarrow A$ such that the composite $A/M \rightarrow A \rightarrow A/M$ is the identity map (first map is hypothetical embedding, second is reduction mod $M$). Then any $A/M$-linear map $M/M^2 \rightarrow A/M$ can be turned into an $A$-linear derivation $A \rightarrow A/M$. The essential feature we have on manifolds is that for the local ring $A$ of smooth functions at any point, the residue field $A/M$ naturally lives inside $A$. This is because the residue field of the local ring at any point is isomorphic to ${\mathbf R}$, and we can naturally think of ${\mathbf R}$ inside the local ring at the point as the constant functions at that point.

The assumption that the residue field of a local ring has an embedding back into the local ring *and* this embedding is a section to the natural reduction homomorphism $A \rightarrow A/M$ could be said in a more down-to-earth way: there is a subfield of $A$ that is a set of representatives for the residue field $A/M$. That is, there's some field $K$ inside $A$ such that the natural map $A \rightarrow A/M$ is onto when we restrict it to $K$. If that's the case, the map $K \rightarrow A/M$ is onto and it's automatically one-to-one since any ring hom. out of a field is one-to-one. Thus $K$ is isom. to $A/M$ so the inverse isomorphism $A/M \rightarrow K$ is a section to $A \rightarrow A/M$.

In the setting of manifolds, $A/M$ is the real numbers and $K$ is always the set of constant functions on a neighborhood of the point of interest, so $K$ is a copy of $A/M$ sitting inside $A$.

first orderat $P$ (and hence to be a directional derivative in local coordinates: $\partial(f) = D_v(f)$ where $v = \sum \partial(x_i) e_i$), as a directional derivative should. Indeed, if $f$ vanishes to first order at $P$ then for local coordinates $(x_i)$ with $x_i(P) = 0$, Taylor's theorem (the real content!) implies $f = \sum x_i x_j h_{ij}$ for smooth $h_{ij}$ near $P$, so $\partial(f) = 0$ by Leibnitz. $\endgroup$ – user30180 Dec 28 '12 at 4:02