The finiteness is not necessary. Every simplicial complex (and more generally, every CW complex) is locally connected (more generally, it is locally contractible).
You can find a proof in Allen Hatcher's Algebraic Topology, in the Appendix, which deals with the topology of CW complexes. Each simplicial complex is a CW complex since a $\triangle^n$ is homeomorphic to the $n$-ball $D^n$
Sorry for the bad quality of the picture. It shows a possible triangulation of the Torus.
Whenever you try to find a triangulation (in particular a Euklidian simplicial complex), it's a collection $K$ of simplicies satisfying the following conditions:
1.) $\sigma\in K$ $\Rightarrow$ every face of $\sigma$ is in $K$.
2.) The intersection of any two simplicies is either empty or a face of both.
3.) For every point in $\sigma\in K$ there exists a neighbourhood which intersects finitely many simplices of $K$.
Let $\{v_0,...,v_n\}$ be the vertices of $\sigma \in K$. A face of $\sigma$ is the simplex spanned by each non-empty subset of $\{v_0,...,v_n\}$.
The most important point is: If you try to triangulate a surface, any two triangles share a single edge or a single vertex or they're disjoint.
A non-example:
This triangulation is false, because the two "triangles" are no triangles. They share 3 edges and the vertices are identified (more on: Triangulation of Torus).
If you would like to calculate the homology goup by using a CW complex, you just take a 0-cell and attach two 1-cells. Then you attach one 2-cell.
The following video gives you an impression how to glue the 2-cell, although this video is not originally made for illustrating this:
https://www.youtube.com/watch?v=nLcr-DWVEto
To get an intuitive understanding of the subject, I recommend you to watch the videos of Wildberger: https://www.youtube.com/watch?v=Uq4dTjHfLpI
Best Answer
There is no general recipe for this, but here is a construction which might help you. Suppose that $X$ is obtained by gluing certain simplices along faces (via affine maps). The result (which is a cell complex) may or may not be a simplicial complex, as you can see by looking at the dunce hat. You can, however, imitate the 2nd barycentric subdivision for $X$ as follows. Subdivide each face of $X$ barycentrically; you obtain a new complex $X'$, which still can fail to be simplicial (this happens in the dunce hat case). However, if you subdivide $X'$ barycentrically again, you obtain a cell complex $X''$ which carries a natural structure of a simplicial complex as intersection of any two faces is either empty or a face. Try it with the dunce hat and see what you get. You can also use this if faces of $X$ are not simplices, but, say, polygons: You first triangulate each polygon and then repeat the 2nd barycentric subdivision construction above. (There are more efficient way, but this one will work.) Try it with the torus obtained by gluing opposite sides of the square in the standard fashion.