First let us get rid of non-compact Riemann surfaces: every holomorphic vector bundle of any rank on such a surface is holomorphically trivial and this seals the fate of its tangent bundle: it is trivial.
From now on, we assume the Riemann surface $X$ is compact.
Here are three useful ways to compute its genus $g$ (which is equivalent to computing the degree of its tangent bundle because of the formula $deg (T_X)=2-2g$ you mentioned) :
I) If the Riemann surface $X\subset \mathbb P^2( \mathbb C)$ is the smooth zero locus of a homogeneous polynomial $f(x_0,x_1,x_2)\in \mathbb C[x_0,x_1,x_2]$ of degree $d$ whose partial derivatives $\frac{\partial f}{\partial x_i}(x_0,x_1,x_2)$ don't vanish simultaneously on $X$ then the genus $g$ of $X$ is
$$ g= \frac{(d-1)(d-2)}{2} $$
II) Suppose the Riemann surface $X\subset \mathbb P^3( \mathbb C)$ is the complete intersection of two surfaces given by the homogeneous polynomials$f(x_0,x_1,x_2,x_3)$ and $g(x_0,x_1,x_2,x_3)$ of degrees $d,e$ whose gradients are linearly independent along $X$. Then $X$ has genus
$$g=1+ \frac{de(d+e-4)}{2} $$
Edit (March 2017)
This formula can be obtained by remembering that the genus of X is equal to its arithmetic genus $p_a=1-P_X(0)$, where $P_X(t)$ is the Hilbert polynomial
of X. For the calculation of that polynomial, see Hassett page 210. See also Hartshorne, Exercise 7.2.(d). page 54.
III) If $\phi:X\to Y$ is a ramified covering of degree $d$ between two compact Riemann surfaces, with $r$ ramification points (counted with multiplicities), then their genera are related by the Riemann-Hurwitz formula :
$$g(X)= 1+ \frac{r}{2} +d\cdot(g(Y)+1) $$
Bibliography
The astonishing result that on a non-compact Riemann surface every holomorphic vector bundle of any rank is holomorphically trivial is proved, with all necessary prerequisites, in the fundamental reference O.Forster, Lectures on Riemann Surfaces , Theorem 30.3.
(Note that if you replace "holomorphic" by "algebraic" the result becomes completely false: already non trivial algebraic line bundles are plentiful on non-complete non-rational algebraic curves)
The Riemann-Hurwitz formula is proved on page 140 of the same book.
For calculations of group actions on a Riemann Surfaces and the associated ramified covering, in relation to Riemann-Hurwitz, I recommend R.Miranda, Algebraic Curves and Riemann Surfaces , Ch.III, ยง3 .
Edit: An example
As an illustration of II), take the quadrics $x_0x_2-x_1^2=0$ and $x_0x_1+x_1x_2-x_3^2=0$ in $\mathbb P^3(\mathbb C)$ . Their intersection in $\mathbb P^3(\mathbb C)$ is a smooth curve of genus $1$ and the last chapter of Harris's Algebraic Geometry will show you that it is isomorphic to the curve $y^2z-x^3-xz^2=0$ in $\mathbb P^2(\mathbb C)$, which indeed has genus $1$ by I).
Best Answer
Riemann surfaces are one dimensional complex manifolds. It is in fact true that any complex $n$-manifold is orientable as a real manifold. One possible way of showing this, is to calculate the determinant of the jacobians of the transition functions.
Suppose you have an $n$-dimensional complex manifold $M$. That is, $M$ is a (first countable, Hausdorff) space such that every point $p\in M$ has an open neighborhood $V$ homeomorphic to an open subset $\Omega\subset\mathbb C^n$, $$\phi:V\rightarrow\Omega~\mathrm{homeomorphism}$$ and when two such neighborhoods intersect, the transition functions are holomorphic, $$\psi\circ\phi^{-1}:\phi(V\cap V')\rightarrow\psi(V\cap V')~\mathrm{biholomorphism.}$$ By identifying $\mathbb C^n$ with $\mathbb R^n$ like so: $(z^1,\dots,z^n)\leftrightarrow(x^1,y^1,\dots,x^n,y^n)$ where $z^k=x^k+iy^k$ is the real part/imaginary part decomposition, we get a real manifold structure on $M$. We can calculate the jacobian matrices of the transition functions for both structures.
Let's set up some notation. The first coordinate chart will be $\phi:V\rightarrow\Omega, p\mapsto (z^1(p),\dots,z^n(p))$, and the second will be $\psi:V'\rightarrow\Omega', p\mapsto (Z^1(p),\dots,Z^n(p)).$ In the complex case, we have $$\mathrm{Jac}_{\phi(p)}(\psi\circ\phi^{-1})=\left(\frac{\partial~[\psi\circ\phi^{-1}]^k}{\partial~z^l}\right)_{1\leq k,l\leq n}=\left(\frac{\partial~Z^k(z^1,\dots,z^n)}{\partial~z^l}\right)_{1\leq k,l\leq n}\\=\left(c_l^k\right)_{1\leq k,l\leq n}\in\mathrm{GL}(n,\mathbb C),$$ and in the real case you'll find $$\mathrm{Jac}_{\phi(p)}(\psi^{\mathbb R}\circ(\phi^{\mathbb R})^{-1})= \left(\begin{array}{cc} \frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l} \\ \frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l} \end{array}\right)_{1\leq k,l\leq n}= \left(\begin{array}{cc} \mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\ \mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k) \end{array}\right)_{1\leq k,l\leq n}\in\mathrm{GL}(2n,\mathbb R),$$ using the Cauchy-Riemann equations. We will calculate the determinant of these matrices, show that it is always $>0$, which is equivalent to $M^{\mathbb R}$ being orientable.
We move on to calculating the determinants of these. Consider the $\mathbb R$-algebra homomorphism $$\rho:M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R),~\left(c_l^k\right)_{1\leq k,l\leq n}\mapsto \left(\begin{array}{cc} \mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\ \mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k) \end{array}\right)_{1\leq k,l\leq n}$$ Since it is $\mathbb R$-linear and the spaces involved are finite dimensional, it is continuous. Also, being an algebra homomorphism, we have $\mathrm{det}~\rho(P^{-1}AP)=\mathrm{det}(\rho(P)^{-1}\rho(A)\rho(P))=\mathrm{det}(\rho(A))$. Finally, the diagonalizable matrices are dense in $M_n(\mathbb C)$, so we can restrict our calculations to diagonal matrices in $M_n(\mathbb C)$. For these, the calcuations are easy: $$\mathrm{det}\big(\rho(\mathrm{Diag}(c_1,\dots,c_n))\big)=\mathrm{det}~\mathrm{Diag}\left(\left(\begin{array}{cc} \mathrm{Re}(c_1) & -\mathrm{Im}(c_1) \\ \mathrm{Im}(c_1) & \mathrm{Re}(c_1) \end{array}\right),\dots,\left(\begin{array}{cc} \mathrm{Re}(c_n) & -\mathrm{Im}(c_n) \\ \mathrm{Im}(c_n) & \mathrm{Re}(c_n) \end{array}\right)\right)\\=\prod_{i=1}^n\mathrm{det}\left(\begin{array}{cc} \mathrm{Re}(c_i) & -\mathrm{Im}(c_i) \\ \mathrm{Im}(c_i) & \mathrm{Re}(c_i) \end{array}\right)=\prod_{i=1}^n|c_i|^2=\left|\mathrm{det}~\mathrm{Diag}(c_1,\dots,c_n)\right|^2,$$ so we conclude that $$\forall A\in M_n(\mathbb C),~\mathrm{det}~\rho(A)=|\mathrm{det}~A|^2$$
Finally, we can conclude that the transition functions for the charts $\phi^{\mathbb R}$ for $M^{\mathbb R}$ have positive determinants, thus the real underlying manifold $M^{\mathbb R}$ is orientable.