I think Hunter and Greg's answers make it hard to see the forest for the trees. Let X be a compact Riem. surface of genus >= g. Let Y be the universal cover of X equipped with the complex structure pulled back from X. As a complex manifold, Y is isomorphic to the upper half plane, and the deck transformations form a subgroup Gamma of PSL_2(R). There will be characters chi of Gamma for which there are nonzero functions f on Y such that f(gz) = chi(g) f(z). For chi ample enough (not defined here), we will be able to choose functions (f_1, f_2, f_3) such that z --> (f_1(z) : f_2(z) : f_3(z)) gives an immersion X --> P^2. All of this works in any genus.
The technical issue is that this map is an immersion, not an injection, meaning that the image can pass through itself. One can either decide to live with this, or work with maps to P^3 instead.
Most books that I have seen don't lift all the way to the universal cover of X. Instead, they take the covering of X which corresponds to the commutator subgroup of pi_1(X). This can be motivated in a particular nice way in terms of the Jacobian. This is a complex manifold with the topological structure of a 2g dimensional torus. There is a map X --> J, so that the map pi_1(X) --> pi_1(J) is precisely the map from pi_1(X) to its abelianization. People then work with the universal cover of J, and the preimage of X inside it. This has three advantages: the universal cover of J is C^g, not the upper half plane; the group of Deck transformations is Z^{2g}, not the fundamental group of a surface, and the action on C^g is by traslations, not Mobius transformations. The functions which transform by characters, in this setting, are called Theta functions*, and they are given by explicit Fourier series.
*This is a slight lie. Theta functions come from a certain central extension of the group of Deck transformations. It is certain ratios of Theta functions that will transform by characters as sketched above. The P function itself, for example, is a ratio of four Theta functions. In the higher genus case, in my limited reading, I haven't seen names for these ratios, only for the Theta functions.
Let's say that your (compact!) Riemann surface is $X$ (of genus at least $1$). What does it mean to say that $J^{g-1}$ is the "variety of degree $g-1$ line bundles?" One way to formalize this is to say that there is a line bundle $\mathcal{L}$ on the product $X \times J^{g-1}$ with the property the rule $p \mapsto \mathcal{L}|_{X \times \{ p \}}$ defines a bijection between the ($\mathbb{C}$-valued) points of $J^{g-1}$ and the line bundles of degree $g-1$. (A stronger statement is that $J^{g-1}$ represents the Picard functor, but I'm going to try to sweep this under the run.)
Let $\pi \colon X \times J^{g-1} \to J^{g-1}$ denote the projection maps. Recall that the formation of the direct image $\pi_{*}(\mathcal{L})$ does not always commute with passing to a fiber. However, the theory of cohomology and base change describes how the fiber-wise cohomology of $\mathcal{L}$ varies. The main theorem states that there is a 2-term complex of vector bundles
$d \colon \mathcal{E}_0 \to \mathcal{E}_1$
that computes the cohomology of $\mathcal{L}$ universally. That is, for all morphisms $T \to J^{g-1}$, we have
$\operatorname{ker}(d_{T}) = (\pi_{T})_{*}(\mathcal{L}_{X \times T})$
and
$\operatorname{cok}(d_{T}) = (R^{1}\pi_{T})_{*}(\mathcal{L}_{X \times T}).$
(The most important case is where $T = \operatorname{Spec}(\mathbb{C})$ and
$T \to J^{g-1}$ is the inclusion of a point.)
The complex $\mathcal{E}_{\cdot}$ is not unique, but any other complex of vector bundles with this property must be quasi-isomorphic.
In the literature, many authors construct a complex $\mathcal{E}_{\cdot}$ using an explicit procedure, but the existence is a very general theorem. I learned about this topic from Illusie's article "Grothendieck's existence theorem in formal geometry," which states the base change theorems in great generality.
In your question, you described how to convert the complex into a line bundle (take the difference of top exterior powers). But now we must check that two quasi-isomorphic complexes have isomorphic determinant line bundles. This can be checked by hand, but this statement has been proven in great generality by Mumford and Knudsen (see MR1914072 or MR0437541). The determinant of the complex $\mathcal{E}_{\cdot}$ is exactly the line bundle you are asking about.
One subtle issue is that the universal line bundle $\mathcal{L}$ is NOT uniquely determined. If $\mathcal{M}$ is any line bundle on $J^{g-1}$, then
$\mathcal{L} \otimes \pi^{-1}(\mathcal{M})$ also parameterizes the degree $g-1$ line bundles in the sense described above (and in a stronger sense that I am sweeping under the rug). However, one can show that $\mathcal{L}$ and $\mathcal{L} \otimes p^{*}(\mathcal{M})$ have isomorphic determinants of cohomology by using the fact that a degree $g-1$ line bundle has Euler characteristic equal to zero.
Added: Here is one method of constructing the complex. Fix a large collection of points $\{x_1, \dots, x_n\} \subset X$ (at least $g+1$ points will do), and let $\Sigma \subset X \times J^{g-1}$ denote the subset consisting of pairs $(x_i,p) \in X \times J^{g-1}$. The universal line bundle $\mathcal{L}$ fits into a short exact sequence:
$$
0 \to \mathcal{L}(-\Sigma) \to \mathcal{L} \to \mathcal{L}|_{\Sigma} \to 0
$$
Here $\mathcal{L}(-\Sigma)$ is the subsheaf of local sections vanishing along $\Sigma$ and $\mathcal{L}|_{\Sigma}$ is the pullback of $\mathcal{L}$ to $\Sigma$.
Associated to the short exact sequence on $X \times J^{g-1}$ is a long exact sequence relating the higher direct images under $\pi$. The first connecting map is a homomorphism
$d : \pi_{*}(\mathcal{L}|_{\Sigma}) \to R^{1}\pi(\mathcal{L}(-\Sigma))$
This is a complex $K_{\cdot}$ of vector bundles that compute the cohomology of $\mathcal{L}$ universally in the sense described earlier. Thus, its determinant is the desired line bundle.
Why is this true? There are two statements to check: that the direct images are actually vector bundles and that the complex actually computes the cohomology of $\mathcal{L}$. Both facts are consequences of Grauert's theorem. Indeed, the hypothesis to Grauert's theorem is that the dimension of the cohomology of a fiber $\pi^{-1}(p)$ is constant as a function of $p \in J^{g-1}$. Using Riemann-Roch (and the fact that we chose a large number of points), this is easily checked. The theorem allows us to conclude that both $\pi_{*}(\mathcal{L}|_{\Sigma})$ and $R^1 \pi(\mathcal{L}(-\Sigma))$ are vector bundles whose formation commutes with base change by a morphism $T \to J^{g-1}$.
An inspection of the relevant long exact sequence shows that the cohomology of the complex
is $\pi_{*}(\mathcal{L})$ and $R^1\pi(\mathcal{L})$ (i.e. the complex compute the cohomology of $\mathcal{L}$). We would like to assert that this property persists if we base-change by a morphism $T \to J^{g-1}$ and work with the complex
$K_{\cdot} \otimes \mathcal{O}_{T}$.
But we already observed that the formation of the direct images appearing in the complex commutes with base-change, and so the base-changed complex fits into a natural long exact sequence. Again, an inspection of this sequence shows that base-changed complex compute the cohomology of $\mathcal{L}|_{X \times T}$, and so the original complex computes the cohomology universally.
Best Answer
Yes, this argument can be made rigorous. One needs three steps.
Step 1. Show that there is at least one smooth plane curve of degree $d$ with the expected genus. Essentially, the proof is given by your heuristic topological argument (deform the union of $d$ lines in general position).
Step 2. Show that if one slightly perturbs the coefficients of a homogeneous polynomial defining a smooth curve, the genus remain unchanged. This is basically a continuity argument.
Step 3. Show that the space $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ of homogeneous polynomials of degree $d$ in three variables defining smooth curves is path-connected. This is because the complement of $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ in $\mathbb{C}[x,\,y,\,z]_d$ (the so-called "discriminant locus") has real codimension $2$.
Putting these three steps together one easily obtains the desired result. For further details you can look at Chapter 4 of Kirvan's book Plane algebraic curves.