You need to be careful what kind of $\mathrm{Pic}$ you are talking about!
The point is that if $X/\mathbb{R}$ is a finite type $\mathbb{C}$-scheme then $X(\mathbb{R})$ (resp. $X(\mathbb{C})$) is a real (resp. complex) manifold (the former being only 'locally a manifold' depending on what sort of axioms of a point-set topological nature you impose). This then allows you to define maps
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))\qquad (1)$$
and maps
$$\mathrm{Pic}_\mathrm{alg}(X_\mathbb{C})\to \mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\qquad (2)$$
But, in general, these maps needn't all be isomorphisms!
For example if $M$ is a smooth real manifold then there is actually an isomorphism
$$\mathrm{Pic}_\mathrm{smooth}(M)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(M)\cong H^1_\mathrm{sing}(M,\mathbb{Z}/2\mathbb{Z})$$
The latter isomorphism comes from the fact that the classifying space of continuous real line bundles is $\mathbb{RP}^\infty$ which is a $K(\mathbb{Z}/2\mathbb{Z},1)$. The former isomorphism can be thought about in two ways:
- The fact that we have smooth approximation for maps $M\to \mathbb{RP}^\infty$
- The fact that smooth real line bundles should be classified by $\mathcal{O}_M^\times$. There is then a SES
$$0\to \mathcal{O}_M\to \mathcal{O}_M^\times\to \underline{\mathbb{Z}/2\mathbb{Z}}\to 0$$
of sheaves and the fact that $H^1(M,\mathcal{O}_M)=H^2(M,\mathcal{O}_M)=0$ because the sheaves $\mathcal{O}_M$ are so-called 'fine' (and so are acyclic).
From this we see that we can refine $(1)$ to
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))$$
but this former map needn't be an isomorphism. As you pointed out, if $X=\mathbb{P}^1_\mathbb{R}$ then this fomer map is
$$\mathrm{Pic}_\mathrm{alg.}(\mathbb{P}^1_\mathbb{R})\cong \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}= \mathrm{Pic}_{\mathrm{smooth}}(\mathbb{RP}^1)$$
where explicitly the map takes $\mathcal{O}(n)$ to the trivial bundle if $n$ is even and the Mobius bundle if $n$ is odd! The point being that while the algebraic structures on $\mathcal{O}$ and $\mathcal{O}(2n)$ (as well as $\mathcal{O}(1)$ and $\mathcal{O}(2n+1)$) are not algebraically equivalent, they are smoothly equivalently. Try it for yourself with overlap maps $x$ and $x^3$!
In the $\mathbb{C}$-case if you assume that $X$ is, in addition, proper then you actually get an isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(X)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))$$
this follows from Serre's GAGA results. Moreover, we can describe the map $\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$ quite nicely. Namely, the complex continuous line bundles on $X(\mathbb{C})$ can be described as $H^2_\mathrm{sing}(X,\mathbb{C})$. Again, the reason for this is that the classifying space of complex line bundles is $\mathbb{CP}^\infty$ which is a $K(\mathbb{Z},2)$. The holomorphic line bundles on $X(\mathbb{C})$ are classified by $H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times)$. The connection between the two is given by the exponential sequence
$$0\to \underline{\mathbb{Z}}\to \mathcal{O}_{X(\mathbb{C})}\to\mathcal{O}_{X(\mathbb{C})}^\times\to 0$$
where the second map is $f\mapsto \exp(2\pi i f)$. Taking the long exact sequence in cohomology we get (part of the ) long exact exponential sequence
$$H^1_\mathrm{sing}(X,\mathbb{Z})\to H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\to H^1(X(\mathbb{C}),\mathcal{O}_X^\times)\to H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\to H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})$$
And, in fact, the natural diagram
$$\begin{matrix}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C})) & \to & \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C})\\ \downarrow & & \downarrow\\ H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times) & \to & H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\end{matrix}$$
commutes with the vertical maps being isomorphisms. This map is called the Chern class of an algebraic/holomorphic line bundle.
So, if $X$ is a smooth projective (geometrically) connected curve of genus $g$ we see that the map
$$\mathrm{Pic}_{\mathrm{alg.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$$
is surjective with kernel a quotient of a vector space of dimension $g$. So, if $X=\mathbb{P}^1_\mathbb{R}$ the kernel is trivial and we get the desired isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(\mathbb{P}^1_\mathbb{C})\cong \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\cong H^2(\mathbb{CP}^1,\mathbb{Z})\cong \mathbb{Z}$$
In fact, what's generally true is that if $X/\mathbb{R}$ is a smooth projective (geometrically) connected curve then the (portion of )the long exact exponental sequence can be written
$$\begin{matrix}H^1_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z}) & \to & H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}) & \to & \mathrm{Pic}(X_\mathbb{C}) & \to & H^2(X(\mathbb{C}),\mathbb{Z}) & \to & H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\\ \downarrow & & \downarrow & & & &\downarrow & & \downarrow\\ \mathbb{Z}^{2g} & &\mathbb{C}^g & & & & \mathbb{Z} & & 0\end{matrix}$$
And, if you take it on faith (this is the beginning of Hodge theory!) that $\mathbb{Z}^{2g}$ is embedded into $\mathbb{C}^g$ as a lattice, then we see that we get a short exact sequence
$$0\to \mathbb{C}^g/\mathbb{Z}^{2g}\to \mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}\to 0$$
so that $\mathrm{Pic}(X_\mathbb{C})$ looks like a disconnected complex Lie group with component group $\mathbb{Z}$ and identity component an abelian variety (i.e. a compact complex Lie group). This identity component is called the Jacobian of $X_\mathbb{C}$ and is denoted $\mathrm{Jac}(X_\mathbb{C})$. The map from
$$\mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}=\pi_0(\mathrm{Pic}(X_\mathbb{C}))$$
is just the degree map. Of course, since $\mathbb{Z}$ is projective and discrete this sequence non-canonically splits to give you that $\mathrm{Pic}(X_\mathbb{C})\cong \mathrm{Jac}(X_\mathbb{C})\times\mathbb{Z}$.
For example, if you take $X=E$ an elliptic curve, then it turns out that $\mathrm{Jac}(E_\mathbb{C})\cong E_\mathbb{C}$!
All of this then starts the fascinating journey into the Albanese variety/Picard scheme of a smooth proper scheme!
The last thing I'll say that is that, in some sense, the algebraic bundles on $X/\mathbb{R}$ smooth projective are much more closely related to the holomorphic bundles on $X(\mathbb{C})$ than the continuous bundles on $X(\mathbb{R})$! In fact, there's the so-called 'Picard-Brauer sequence' which contains the terms
$$0\to \mathrm{Pic}(X)\to \mathrm{Pic}(X_\mathbb{C})^{\mathrm{Gal}(\mathbb{C}/\mathbb{R})}\to \mathrm{Br}(\mathbb{R})(\cong\mathbb{Z}/2\mathbb{Z})$$
From this we see that the algebraic bundles on $X$ embed into the algebraic bundles on $X_\mathbb{C}$ (which is equal to the holomorphic bundles on $X(\mathbb{C})$) and that up to a $\mathbb{Z}/2\mathbb{Z}$-term they exactly hit the Galois invariants of the algebraic bundles on $X_\mathbb{C}$.
In the case of $X=\mathbb{P}^1_\mathbb{R}$ this sequence is not very interesting. It looks like
$$0\to \mathbb{Z}\to\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$
where the map $\mathbb{Z}\to\mathbb{Z}$ is an isomorphism and the map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ is trivial.
But, if instead of $\mathbb{P}^1_\mathbb{R}$ you took it's only non-trivial twist $X:=V(x^2+y^2+z^2)\subseteq\mathbb{P}^2_\mathbb{R}$ then your sequence actually looks like
$$0\to 2\mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$$
The point being that if you take any degree $2$-point on $X$ then $\mathrm{Pic}(X)\cong \{\mathcal{O}(np):n\in\mathbb{Z}\}$. But, since $p$ is a point of degree $2$ when you base change to $\mathbb{C}$ you get that $p$ splits into two points--$q_0$ and its Galois conjugate $\sigma(q_0)$ so that $\mathcal{O}(p)$ maps to $\mathcal{O}(q_0)\otimes \mathcal{O}(\sigma(q_0))\cong \mathcal{O}(2)$.
Best Answer
I'm not sure if this is exactly what you're looking for, but the equivalence of $y^2$ and $(1 - x)(1 + x)$ for $\mathbf{S}^1$ says that you can look at the divisor $D = 2((1,0) + (-1,0))$ in two ways:
By intersecting the lines $x = \pm 1$ with the circle (the lines are tangent so the intersection has multiplicity $2$)
By intersecting the line $y = 0$ with the circle. The points above are doubled so really we want the doubled line $y^2 = 0$.
So we have $D = \mathcal{V}(y^2) = \mathcal{V}((1-x)(1+x))$.
Also notice that every principal divisor $\mathcal{V}(f)$ on $\mathbf{S}^1$ has even degree. For example, every line intersects the circle in two distinct points or one doubled point (a tangential intersection). This classifies the principal divisors (i.e. every even degree divisor is principal).
So ${\rm Cl}(\mathbf S^1) = \mathbf Z/2$. The problem here, is somehow we want to be able to break our divisor $D$ up into 4 parts, not just two. That is, if $(1,0) = \mathcal{V}(f)$ and $(-1,0) = \mathcal{V}(g)$ then we could say
$$ \mathcal{V}(y) = \mathcal{V}(fg), \mathcal{V}(1-x) = \mathcal{V}(f^2), \mathcal{V}(1 + x) = \mathcal{V}(g^2) $$
Which would give us $y^2 = (fg)^2$ and $(1 - x)(1 + x) = (f^2)(g^2)$ (up to a constant) so that $y^2$ and $(1 - x)(1 + x)$ represent the same factorization. But $(1,0)$ and $(-1,0)$ are not principal divisors so we don't get this.