$k[X_1,...,X_{m}]\otimes_kk[X_{m+1},...,X_{n+m}]\simeq k[X_1,...,X_{n+m}]$ 
The ring $k[X_1,...,X_m,X_{m+1},...,X_{n+m}]$, together with the obvious inclusions $$k[X_1,...,X_{m}]\hookrightarrow k[X_1,...,X_{n+m}]\hookleftarrow k[X_{m+1},...,X_{n+m}]$$
is the tensor product of the $k$-algebras $k[X_1,...,X_m]$ and $k[X_{m+1},...,X_{n+m}]$ where $k$ is a commutative ring with unity.

First I want to show this by showing the above two inclusions have a universal property but I don't know how to define the unique $k$-algebra homomorphism $\Phi$ in the following diagram:

where $\phi$ and $\psi$ are the given $k$-algebra homomorphism. How can I define it?
Second, the proof of this statement in Milne's textbook is the following: To verify this we only have to check that, for every $k$-algebra $R$, the map $$\text{Hom}(k[X_1,...,X_{m+n}],R)\to \text{Hom}(k[X_1,...],R)\times\text{Hom}(k[X_{m+1},...],R)\tag{1}$$
induced by the inclusion is a bijection. But this map can be identified with the bijection $$R^{n+m}\to R^m\times R^n.\tag{2}$$
I understand that showing $(1)$ is same as showing universal property but how can $(1)$ be identified with $(2)$? Could you explain this to me?
Milne's book and the statement is in p.42 Example 10.4
Edit: Universal property of polynomial ring in Computational commutative algebra 1 by Martin & Lorenzo

 A: I believe that Milne's book wants you to verify that $k[X_1, \dots, X_{n+m}]$ satisfies the universal property of the coproduct of $k[X_1, \dots, X_n]$ and $k[X_{n+1}, \dots, X_{n+m}]$ in the category of commutative unital rings (which is the tensor product of rings).
Note that $\Phi$ is the unique ring morphism satisfying $\Phi i = \phi $ and $\Phi j = \psi$. In particular, $\Phi(X_i) = \phi(X_i)$ for $i=1,\dots, n$ and $\Phi(X_i) = \psi(X_i)$ for $i = n+1, \dots, n+m$. By the universal property of the polynomial algebra, there is a unique morphism $\Phi: k[X_1, \dots, X_{n+m}]\to R$ satisfying this and you are done.
This universal property (of the polynomial algebra) asserts that if $R$ is a commutative algebra, then we have a bijection
$$\text{Hom}_{Alg}(k[X_1, \dots, X_n], R) \cong R^n: f \mapsto (f(X_1), \dots, f(X_n))$$  I.e., an algebra morphism $k[X_1, \dots, X_n]\to R$ is uniquely determined by the choice of $n$ points of the ring $R$. This explains why $(1)$ and $(2)$ are equivalent.
