First, let us fix some definitions.
Definition: an $n$-simplex is a $n$-dimensional polytope which is the convex hull of it's $n+1$ vertices. Importantly, no vertex is contained in the convex hull of any other vertices.
Definition: a simplicial complex $K$ is a set of simplicies such that any face of a simplex of $K$ is also in $K$ and the intersection of any two simplicies $\sigma_1, \sigma_2$ is a face of $\sigma_1$ and $\sigma_2$.
Part (a): In order to show that these vertices actually form the vertices of simplex, we must check that no vertex is contained in the convex hull of any of the others. Suppose that $b_k=(x_{0,k},x_{1,k},\cdots,x_{n,k})$ (in $v_i$ coordinates) is contained in the convex hull of the other $b_j$. This would mean that there's a nontrivial linear dependence relation on the the set $\{b_j\}$.
But if we have some linear relation $\sum_{i=0}^n a_ib_i=0$, we can replace the $b_i$ by their defining linear combinations $\frac{1}{n+1}(v_0+\cdots+v_n)$ and obtain a linear relation on the $v_i$. But this is clearly ridiculous as the vectors $v_i$ are linearly independent since we assumed they were the vertices of a simplex. Some quick calculation shows that if the coefficients of $v_i$ are all zero, then so too are the coefficients $a_i$.
Part (b): For this, we need to check that the union of these simplicies is the entire space and that any two simplicies intersect in simplices.
To do the first, I'll give a method for determining which simplex a given point lies in. Represent an arbitrary point in our simplex as the linear combination of $v_i$, ie $x=a_0v_0+\cdots+a_nv_n$. I claim that it lies in the simplex determined by $b_{\sigma_0},\cdots,b_{\sigma_n}$ where $\sigma_i$ is the simplex defined by taking the greatest $i+1$ elements of the set $\{v_j\}$ given an order by $v_i\geq v_j$ if $a_i\geq a_j$ and ties broken by the lexographic ordering. Note that this always returns an answer- so every point in our original simplex is in at least one simplex in the barycentric subdivision.
Next, we need to show that if any collection of simplicies intersect, they intersect in a simplex. Since a simplex is determined entirely by its vertices, this means that the intersection set is determined by intersection of the vertices of the simplicies in common. But this means that the intersection of simplicies is exactly a simplex with vertices which are the intersection of the vertex sets of the simplicies we're intersecting. (Since these form a subset of the vertices of a simplex, they again determine a simplex.)
Part (d): Apply (b) to see that the barycentric subdivision forms a simplicial complex. The geometric realizations are the same because the behavior on the barycentric subdivision is still totally determined by the behavior of the points that were inherited from the original simplex. This is a moral-of-the-story answer, because I'm not sure what definition of geometric realization you're working with (the one I know is defined as a functor from simplicial sets to compactly generated hausdorff topological spaces- if you post your definition and you're still interested, I can add more to this.)
Finally, I've noticed from your recent questions that you seem to be in the midst of self-studying a lot of the foundations for algebraic topology. I highly recommend actually taking a course in the subject and discussing issues with your professor and coursemates- live interaction with peers and teachers is a much more reliable and useful tool than consulting stackexchange every time you have a potential issue.
As Eric Wofsey says, there are two notions here: An orientation of an $n$-dimensional simplicial complex is a choice of orientation for each $n$-dimensional simplex. The easiest way to find one is to take a global ordering of all the vertices. If your simplicial complex is a manifold then your ordering will give an orientation of the manifold if and only if, whenever we have two simplices $(u,w_1, w_2, \dots, w_n)$ and $(v,w_1, w_2, \dots, w_n)$ which overlap on an $n-1$ dimensional face, then $(u,w_1, w_2, \dots, w_n)$ and $(v,w_1, w_2, \dots, w_n)$ are given opposite signs. Some manifolds are orientable and some are not. For example, I can trangulate the Mobius strip as the simplicial complex $(v_1, v_2, v_3)$, $(v_2, v_3, v_4)$, $(v_3, v_4, v_5)$, $(v_4, v_5, v_1)$, $(v_5, v_1, v_2)$ and we cannot choose the orientations of these simplices compatibly.
What I want to add to Eric's answer is that, if you have a specific simplicial complex, it is straightforward to work out if we can choose compatible orientations. Let $\Gamma$ be the graph whose vertices are the $n$-dimensional simplices of your complex and where there is an edge between vertices corresponding to adjacent vertices. Find a spanning tree $T$ of this graph. (Or a forest if it is disconnected.) Choose the orientation of one simplex arbitrarily. (If $\Gamma$ is disconnected, choose one vertex in each tree of the spanning forest.) Travel along the edges of $T$; there is a unique way to choose the orientation at each new vertex of $T$ so that the orientations are compatible along the edge of $T$ we have just traveled.
After you have oriented all the simplices to be compatible with the edges in $T$, check if they are also compatible with the other edges of $\Gamma$. If so, you have built an orientation of the manifold, if not the manifold is not orientable.
If, as you say, your simplicial complex is homeomorphic to $S^4$, then you will succeed, because $S^4$ is orientable.
Best Answer
Yes, every simplicial complex homeomorphic to a sphere is orientable. To prove this in the generality that you ask requires some hard work. Perhaps the most modern approach would be first to develop the topological theory of orientation as is done in most algebraic topology courses (such as Hatcher's book; for a more succinct and still very clear description I also like the appendix to Milnor's book). Then apply that theory to spheres and to simplexes.
And yes, for every orientable pseudo-manifold its first barycentric subdivision is orientable. This is quite elementary, you simply have to construct, by induction on dimension, a canonical way to define an orientation on the first barycentric subdivision of an oriented $n$-simplex that agrees inductively with the first barycentric subdivision of its boundary. If you've ever seen a proof of the excision theorem in the theory of singular homology then you've probably encountered this construction.