$K_0$ of a non-separated scheme This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($m\geq 1$). 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 $H\cong \mathbf{P}^{n-1}$ be a hyperplane with complement $U\cong \mathbf{A}^n$. 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 $i:H\longrightarrow X$ 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 $\pi:X\longrightarrow H$ such that $\pi\circ i = \textrm{id}_{H}$. 
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 ($m\geq 1$). 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 $H\cong \mathbf{P}^{n-1,m}$ and that its complement is $\mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:H\longrightarrow \mathbf{P}^{n,m}$ 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, $K_0(\mathbf{P}^{n,m}) \cong K_0(S)^{n+m}$.
 A: This is exactly the sort of example where it is relevant which version of K-theory you are employing. (In particular, which theorems you can employ here depends on this!) The issue here is a tad subtle.
To illustrate, let's restrict attention to the case of affine $n$-space $X$ with a doubled origin ($n\geq 2$). (An iteration of the discussion below, combined with your localization argument in Idea2, will let you address your general multiply-origined projective spaces.) For convenience, let me assume that the base is a regular noetherian scheme $S$.
Let me first answer your question as it was asked. Quillen's localization sequence gives fiber sequences of spectra or spaces
$$G(S)\to G(\mathbf{A}_S^n)\to G(\mathbf{A}_S^n-\{0\})$$
and
$$G(S)\to G(X)\to G(\mathbf{A}_S^n)$$
Here I'm using $G$-theory is the $K$-theory spectrum (or space) of coherent sheaves. This has the property that $G(S)\simeq G(\mathbf{A}_S^n)$. Putting this together, we see that
$$G(X)\simeq G(\mathbf{A}_S^n)\times G(S)\simeq G(S)\times G(S)$$
This answers your question for $G$-theory; in particular $\pi_0$ of $G$ is the Grothendieck group you seek, and so we conclude that $G_0(X)=G_0(S)\oplus G_0(S)$. Now use your localization arguments to get that $G_0(\mathbf{P}^{n,2})$ is $n+2$ copies of $G_0(S)$.
However: for $K$-theory, the way this computation gets done depends critically on which model of $K$ you use.
(A) If we define $K$-theory as in Thomason-Trobaugh (as the Waldhausen $K$-theory spectrum of perfect complexes), then in this case $K(X)\simeq G(X)$ and $K(S)\simeq G(S)$. (This follows from "Poincaré Duality;" see the end of section 3 of Thomason-Trobaugh.)
(B) If, however, we define $K^{\mathrm{naive}}(X)$ as the Quillen $K$-theory of the category of algebraic vector bundles on $X$, then things look different. [N.B. that this is the name given by Grothendieck, Illusie, et al. in SGA 6. It is not meant to be insulting!] Indeed, the inclusion $\mathbf{A}_S^n\to X$ induces an equivalence between the catgories of algebraic vector bundles on $X$ and those on $\mathbf{A}_S^n$, since the origin has codimension at least $2$. So then
$$K^{\mathrm{naive}}(X)\simeq K^{\mathrm{naive}}(\mathbf{A}_S^n)\simeq K(\mathbf{A}_S^n)\simeq K(S).$$
The second equivalence here follows from the fact that "naive" $K$-theory agrees with $K$-theory for schemes that admit an ample family of line bundles (see section 3 of Thomason-Trobaugh). The difference between (A) and (B) here reflects the failure of our $X$ to admit such a family.
An example like this is discussed is Thomason-Trobaugh I think at the end of section 8 (wherever they talk about Mayer-Vietoris). [Sorry, this is from memory; I don't have a copy of the Festschrift handy.]
Hope this helps!
