Dual of the multiplication map $m: A \otimes_k A \rightarrow A$, where $A$ is a fin. dim. algebra over a field $k$. Let $A$ be a finite dim. $k$-algebra. Let $m:A \otimes_k A \rightarrow A$ be the multiplication map of $A$. Take the dual of this map as follow
$$\delta:=\operatorname{D}(m) : \operatorname{D}(A) \rightarrow \operatorname{D}(A)\otimes_k \operatorname{D}(A),$$
where $\operatorname{D} = \operatorname{Hom}_k (-,k)$. 
How is this $\delta$ defined explicitly by using $m$?
 A: The elements of $D(A)$ are maps $f:A \to k$, and the elements of $D(A) \otimes D(A)$ are maps of the form 
$$
(a,b) \mapsto \sum_i f_i(a)g_i(b), \quad \text{for all } a,b \in A
$$ 
for some $f_i,g_i \in D(A)$.  That is, $D(A) \otimes D(A)$ is the space of all bilinear maps over $A$.
With that: for any $f \in D(A)$, we have
$$
(\delta(f))(a,b) = (f \circ m)(a,b) = f(ab).
$$
So, $\delta$ maps $f \in D(A)$ to the bilinear map $\delta(f): (a,b) \mapsto f(ab)$.

Strictly speaking, if we consider $m$ to be a linear function on $A \otimes A$ and we apply the duality functor $D$, then we should find that the target space of $\delta$ should be $D(A \otimes A)$ rather than $D(A) \otimes D(A)$.  With that, we would find that $\delta$ maps $f \in D(A)$ to the unique linear map $\delta(f):A \otimes A \to k$ satisfying $\delta(f): a \otimes b \mapsto f(ab)$.
That being said: because of the universal property that defines tensor products, these two $\delta$'s are the same up to canonical isomorphism.
A: This is really a comment and an addendum on Omnomnomnom's answer, but too long (even longer than the word "Omnomnomnom")... But it's worthwhile, at least just once, to work out a toy example. 
Take the $2$-dimensional algebra $A=  k[x]/(x^2)$, i.e., $A=k[x]$, with $x^2 = 0$. 
Now, since $D(A)$ is finite dimensional, we can identify $D(A)\otimes D(A)$ with $D(A\otimes A).$ Use the obvious basis $1$ and $x$ for $A$, and consider the dual basis $\epsilon$ and $\lambda$, i.e., $$\text{ $\epsilon(1)=1$, and $\epsilon(x)=0$, and  $\lambda(1)=0$ and $\lambda(x)=1$.}$$ Then $A\otimes A$ has basis 
$$\text{$1\otimes 1$, $x\otimes 1$, $1\otimes x$, $x\otimes x$,}$$ and 
$D(A\otimes A)$ has corresponding dual basis 
$$\text{$\epsilon\otimes \epsilon $,  $\lambda \otimes \epsilon$, $\epsilon \otimes \lambda$,$\lambda \otimes \lambda$.}$$ Since, as "O" above writes, $$(\delta f )(\sum a\otimes b) = \sum f(ab),$$ 
by testing $\delta(\epsilon)$ and $\delta ( \lambda)$ on our $A\otimes A$ basis elements,  we see that 
$$\text{ $\delta(\epsilon) = \epsilon\otimes\epsilon$ and $\delta (\lambda)= \lambda\otimes\epsilon + \epsilon\otimes \lambda $.}$$
