Here:
$A$ is a ring.
$E,E'$ are right $A$-modules, $F,F'$ are left $A$-modules.
$u:E\rightarrow E'$ and $v:F\rightarrow F'$ are $A$-module homomorphisms.
With this setup, I didn't understand following paragraph (its last line) from Bourbaki's Algebra 1.
Can one explain it (especially the last line)?
Things might be clearer if we actually use a name for such canonical mapping, say $$\Phi:{\rm Hom}_A(E,E')\otimes_{\Bbb Z}{\rm Hom}_A(F,F')\to {\rm Hom}_{\Bbb Z}(E\otimes_A F,E'\otimes_AF').$$In its full glory, this map satisfies $\Phi(u\otimes v)(x\otimes y) = u(x)\otimes v(y)$, for all $x\in E$ and $y\in F$. Maybe one possible cause for confusion is that $u$ and $v$ are being used to denote linear maps and not elements of a "base" module.
Anyway: $\Phi$ might not be injective, because we can have non-zero maps $u$ and $v$ producing the zero map. For example, take $A=\Bbb Z$, $E'$ to be a divisible abelian group and $F'$ to be a torsion group. So $E'\otimes_{\Bbb Z}F'=0$. But we can take $E=E'$, $F=F'$, and $u$ and $v$ to be the corresponding identity maps. I am giving you a concrete situation and a non-zero element ${\rm Id}_E\otimes {\rm Id}_F$ in the kernel of $\Phi$.
Thing is, one can also cook up an example where $\Phi$ is not surjective (I would have to think harder for this one right now, though).
The bottom line here is that one omits $\Phi$ from the notation, wanting to write $u\otimes v$ again for what I have called $\Phi(u\otimes v)$. But since $\Phi$ is not injective, you cannot identify the element $u\otimes v$ in the domain of $\Phi$ with the corresponding linear map $u\otimes v$. To avoid ambiguity, you have to say exactly on which space this lives.
It is somewhat similar to the situation where you have two modules $M_1\subseteq M_2$, and a non-flat module $N$: the inclusion $M_1\hookrightarrow M_2$ may induce a non-injective map $M_1\otimes N\to M_2\otimes N$, meaning that the notation $m_1\otimes n$ is ambiguous for $m_1\in M_1$ and $n\in N$, since it can represent distinct elements in $M_1\otimes N$ and $M_2\otimes N$.