Can the fundamental group and homology of the line with two origins be computed as a direct limit?

Yes, $X$ is the direct limit of this sequence. This is essentially immediate from the universal property of quotient spaces: a map out of the direct limit is just a map out of $\mathbb{R}\sqcup \mathbb{R}$ which coincides on the two copies outside of $(-1/n,1/n)$ for all $n$, and that just means it coincides everywhere except $0$.

You can deduce the homology and homotopy groups of $X$ from the following pair of facts, which immediately imply that $H_*$ and $\pi_*$ preserve the direct limit:

Let $K$ be a compact Hausdorff space. Then:

1. Any continuous map $f:K\to X$ lifts to a continuous map $K\to X_n$ for some $n$.
2. If $g,h:K\to X_m$ are continuous maps whose compositions with $X_m\to X$ are equal, then there is some $n\geq m$ such that their compositions with $X_m\to X_n$ are equal.

To prove (1), let $0$ and $0'$ be the two origins in $X$ and let $A=f^{-1}(\{0\})$ and $A'=f^{-1}(\{0'\})$. Then $A$ and $A'$ are disjoint closed subsets of $K$, so they have disjoint open neighborhoods $U$ and $U'$. Note that then $\partial U$ is closed and disjoint from $A$ and $A'$, and so $f(\partial U)$ is a compact subset of $X\setminus\{0,0'\}$. This means there is some $n$ such that $f(\partial U)\cap [-1/n,1/n]=\emptyset$. Now lift $f$ to $X_n$ by mapping it to the first copy of $\mathbb{R}$ on $U$ and the second copy of $\mathbb{R}$ on $K\setminus U$. This is obviously continuous on $U$ and on the interior of $K\setminus U$. On $\partial U$, it is continuous since the two copies of $\mathbb{R}$ are identified in $X_n$ on a neighborhood of the image of $\partial U$.

To prove (2), let $f:K\to X$ be the common composition of $g$ and $h$ with $X_n\to X$ and let $A=f^{-1}(\{0,0'\})$. Note that then $g$ and $h$ must coincide in a neighborhood $U$ of $A$: if $g(x)=h(x)=0$, then there is a neighborhood of $x$ on which both $g$ and $h$ map into the first copy of $\mathbb{R}$, and so must coincide, and similarly if $g(x)=h(x)=0'$. By compactness of $K\setminus U$, $U$ must contain $f^{-1}([-1/n,1/n])$ for some $n\geq m$. Since $g$ and $h$ coincide on $U$, this means that their compositions with $X_m\to X_n$ are equal.