There are many ways to construct triangle-free graphs with large chromatic number. One well-known construction is Jan Mycielski's, which is described in the Wikipedia article that you cited; I guess you had trouble understanding it. Instead of explaining Mycielski's construction, I will describe another one, which is even less "efficient" (meaning that it produces a bigger graph for a given chromatic number) but maybe slightly simpler.
For a given natural number $n\in\mathbb N$, let $G_n$ be the following graph with $\binom n2$ vertices and $\binom n3$ edges: the vertices are the pairs $(x,y)$ of integers with $1\le x\lt y\le n$; for each triple $(x,y,z)$ with $1\le x\lt y\lt z\le n$, there is an edge joining vertex $(x,y)$ to vertex $(y,z)$. It is plain to see that $G_n$ is triangle-free. I will show that $G_n$ is $k$-colorable if and only if $n\le2^k$. Therefore, if $2^{1525}\lt n\le2^{1526}$, then $\chi(G_n)=1526$.
First, suppose $G_n$ is $k$-colorable; I will show that $n\le2^k$. Let $f:V(G_n)\to[k]=\{1,\dots,k\}$ be a proper vertex-coloring of $G_n$; i.e., for $1\le x\lt y\le n$, the vertex $(x,y)$ gets color $f(x,y)\in[k]$. For each $x\in[n]=\{1,\dots,n\}$, let $S_x=\{f(u,x):1\le u\lt x\}\subseteq[k]$. Note that, since $f$ is a proper coloring, the sets $S_1,\dots,S_n$ must be distinct; namely, if $1\le x\lt y\le n$, then $f(x,y)\in S_y\setminus S_x$, showing that $S_x\ne S_y$. Thus there are $n$ distinct subsets of $[k]$, i.e., $n\le2^k$.
Now, suppose $n\le2^k$; I will show that $G_n$ is $k$-colorable. Let $S_1,\dots,S_n$ be $n$ distinct subsets of $[k]$, indexed in order of increasing size, so that $|S_1|\le|S_2|\le\dots\le|S_n|$; it follows that $S_y\not\subseteq S_x$ whenever $1\le x\lt y\le n$. Finally, we get a proper coloring $f:V(G)\to[k]$ by assigning to each vertex $(x,y)$ a color $f(x,y)\in S_y\setminus S_x$.
It depends on what you consider a plane drawing to be, but how about something like:
Put a vertex at $(0,0)$ and one at $(1,x)$ for every $x\in\mathbb R$, with a straight edge going between it and $(0,0)$.
Naively this seems to satisfy the requirements of a planar graph drawing: No point lies on two edges (except endpoints), no vertex lies on an edge it is not an endpoint of, every edge is represented by a continuous path between its endpoints.
On the other hand: A problem with this is that if we consider an abstract graph a topological space (with one point per vertex and a copy of the unit interval for each edge, stitched together in the obvious way), then the topology of the above drawing as a subspace of $\mathbb R^2$ is not the same as that of the graph.
Indeed if we define a "plane drawing" of a graph to be a subset of $\mathbb R^2$ which (with the subspace topology) is homeomorphic to the graph, then there can't be any uncountable planar graph, simply because there's no uncountable discrete subset of $\mathbb R^2$ that can be the image of the vertices.
On the other other hand: There are some topological subtleties hidden beneath "stitched together in the obvious way" here. Actually, as soon as there's a node with countable degree the most obvious way of stitching together (resulting in a quotient space) gives something that is not homeomorphic to a straightforward drawing of the graph -- such as a node at $(0,0)$ and one at $(1,n)$ for every $n\in\mathbb N$, with a straight edge to $(0,0)$. This can be fixed, though, by definig the stiching in a slightly more ad-hoc way which gives a different topology on the abstract graph.
Best Answer
Just like a planar graph: a graph that can be embedded in the plane such that its vertices are points of the plane and the edges can be "realised" as actual homeomorphic copies of $[0,1]$ (simple curves) that only intersect at vertices but nowhere else. Just as how you draw planar graphs. Now replace the plane by a surface and you have your definition. It says nothing on the distribution of the points (or size etc), but that we can topologically embed the graph (seen as a one-dimensional simplex) into the surface. For nice surfaces (differential or topological $2$-manifolds, really) we still have some analogue of the Jordan curve theorem, in the sense that a graph-cycle bounds some unique "interior", just as in the plane,so the notion of a "face" makes sense for such an embedding.