algebraic k theory - $K_0$ of a non-separated scheme

This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($mgeq 1$). For any (noetherian) scheme $X$, let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$.

Firstly, let me sketch that $K_0(mathbf{P}^n) cong K_0(mathbf{P}^{n-1})oplus K_0(mathbf{A}^n)$.

Let $X=mathbf{P}^n$ be the projective $n$-space.

(I omit writing the base scheme in the subscript. In fact, you can take any noetherian scheme as a base scheme in the following, I think.)

Let $Hcong mathbf{P}^{n-1}$ be a hyperplane with complement $Ucong mathbf{A}^n$. By a well-known theorem on Grothendieck groups, we have a short exact sequence of abelian groups $$K_0(H) rightarrow K_0(X) rightarrow K_0(U) rightarrow 0.$$ Now, let $i:Hlongrightarrow X$ be the closed immersion. Then the first map in the above sequence is given by the "extension by zero", which in this case is just the K-theoretic push-forward $i_!$, or even better, just the direct image functor $i_ast$. Now, there is a projection map $pi:Xlongrightarrow H$ such that $picirc i = textrm{id}_{H}$.

By functoriality of the push-forward, we conclude that $pi_! circ i_ast = pi_! circ i_! = textrm{id}_{K_0(H)}$.

Therefore, we may conclude that $i_ast$ is injective and that we have a split exact sequence $$0 rightarrow K_0(H) rightarrow K_0(X) rightarrow K_0(U) rightarrow 0.$$ Thus, we have that $K_0(mathbf{P}^n) cong K_0(mathbf{P}^{n-1})oplus K_0(mathbf{A}^n)$.

Q1: Let $mathbf{P}^{n,m}$ be the projective $n$-space with $m$ origins ($mgeq 1$). For example, $mathbf{P}^{n,1} = mathbf{P}^n$. (Again the base scheme can be anything, I think.) Now, is it true that $$K_0(mathbf{P}^{n,m}) cong K_0(mathbf{P}^{n-1,m}) oplus K_0(mathbf{A}^n)?$$

Idea1: Take a hyperplane $H$ in $mathbf{P}^{n,m}$. Is it true that $Hcong mathbf{P}^{n-1,m}$ and that its complement is $mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:Hlongrightarrow mathbf{P}^{n,m}$ is proper, right? Also, is the projection $pi:mathbf{P}^{n,m}rightarrow H$ proper? If yes, the above reasoning applies. If no, how can one "fix" the above reasoning? I think that in this case one could still make sense out of $i_!$ and $pi_!$ (even if they are not proper maps.)

Idea2: Maybe it is easier to show that $K_0(mathbf{P}^{n,m}) cong K_0(mathbf{P}^{n-1})oplus K_0(mathbf{A}^{n,m})$, where $mathbf{A}^{n,m}$ is the affine $n$-space with $m$ origins. Then one reduces to computing $K_0(mathbf{A}^{n,m})$...

Idea3: One could also take $m=2$ as a starting case and look at the complement of one of the origins. Then we get a similar exact sequence as above and one could reason from there.

Which of these ideas do not apply and which do?

Note: Suppose that the base scheme is a field. Since $K_0(mathbf{A}^n) cong mathbf{Z}$ and $K_0(mathbf{P}^n) cong mathbf{Z}^{n+1}$, this would show that $$K_0(mathbf{P}^{n,m}) cong mathbf{Z}^{n+m}.$$ More generally, if $S$ is the base scheme, $K_0(mathbf{P}^{n,m}) cong K_0(S)^{n+m}$.