Tensor Product of Extension of Scalars Let $M$ and $N$ be modules over commutative ring $A$. Let $\varphi:A\to B$ be a morphism of rings. We use the notation, $M_B = M\otimes_A B$, this is a module over $A$, but we will rather consider $M_B$ as a module over $B$. We will prove (assuming this is even true) that $(M\otimes_A N)_B = M_B \otimes_B N_B$. In what follows we will use the categorical definition of the tensor product. 
Let $P=M\otimes_A N\otimes_A B$ (which is =$(M\otimes_A N)_B$). Now $P$ is some module/$A$, which satisfies the universal multi-linear property, we use the notation $m\otimes n\otimes b$ to denote the image of this universal multilinear map $M\times N\times B\to P$. We can regard $P$ as a module/$B$, and will think of it in this way from now on. 
Since we claim that $(M\otimes_A N)_B = M_B \otimes_B N_B$, we need to show, that $P$ satisfies the categorical tensor product. Define the map, $\beta:M_B\times N_B\to P$ by $\beta(m\otimes b,n\otimes b') = m\otimes n\otimes bb'$, this map is seen to bilinear/$B$. Now given any bilinear map $f:M_B\times N_B\to L$, where $L$ is a module/$B$, define $g:P\to L$ as, $g(m\otimes n\otimes b) = f(m\otimes 1,n\otimes b)$. This is a linear map/$B$. Furthermore, $f = g\circ \beta$. Thus, the universal property been satisfied. 
i) Why is $\beta$ well-defined? Can this be seen by the property of the tensor product? Keep in mind, $P$ is now regarded as a module/$B$, so whatever we could have said about it as module/$A$, does not follow. 
ii) Why is $f$ and $g$ well-defined? Similar issues as in (i)?  
 A: For $\beta$: Fix $n \in N$,  $b' \in B'$. Define a map $b_{n, b'}: M \times B \to P$ by $b_{n,b'}(m, b) := m \otimes n \otimes bb'$. As $b_{n,b'}$ is $A$-bilinear, we get an induced map $\bar b_{n,b'} \colon M_B \to P$. Note, that
$$ \bar b_{n,b'}\bigl(b''(m \otimes b)\bigr) = b_{n,b'}(m, b''b) = b''(m \otimes n \otimes b) $$
(by definition of the $B$-module action in $P$), so $b_{n,b'}$ is actually $B$-linear. 
Now define a map $\bar b \colon N \times B \to {\rm Hom}_B(M_B, P)$ by $\bar b(n,b') := \bar b_{n,b'}$. $\bar b$ is $A$-bilinear, hence induces a map $\tilde \beta \colon N_B \to {\rm Hom}_B(M_B, P)$. Again, we see by direct computation, that $\tilde \beta$ is actually $B$-linear. As 
$$ {\rm Hom}_B\bigl(N_B, {\rm Hom}_B(M_B, P)\bigr) \cong {\rm Hom}_B(N_B \otimes_B M_B, P) $$
(universal property) and $\beta$ corresponds to $\tilde\beta$ under this isomorphism, $\beta$ is well defined. 
For $f$: $f$ is a given bilinear map, I do not see any problems regarding well-definedness.
For $g$: Define $G \colon M \times N \times B \to L$ by $G(m,n,b) := f(m \otimes 1, n \otimes b)$. Then $G$ is $A$-trilinear, hence induces a map $g \colon P \to L$, with $g(m\otimes n \otimes b) = f(m \otimes 1, n \otimes b)$. It remains to check, that $g$ is $B$-linear, again, direct computation shows:
\begin{align*}
  g\bigl(b' (m \otimes n \otimes b)\bigr)
   &= g(m \otimes n \otimes b'b)\\
   &= f(m \otimes 1, n \otimes b'b)\\
   &= f(m \otimes 1, b'(n\otimes b)\bigr)\\
   &= b'f\bigr(m \otimes 1, n \otimes b) 
\end{align*}
