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
Let $f\colon E\to X$ be any schematically dominant morphism of schemes, e.g., a surjective morphism with $X$ reduced. Let $U_{E/X}$ be the cokernel of the canonical morphism $f^{\flat}\colon\mathbb G_{m,X}\to f_{*}\mathbb G_{m,E}$ of etale sheaves on $X$ (this morphism is injective by the schematically dominant hypothesis. The sheaf just defined is, or should be, called the sheaf of relative units of $E$ over $X$). Then there exists a canonical exact sequence of abelian groups $$ 0\to\varGamma(X, \mathbb G_{m,X})\to\varGamma(E, \mathbb G_{m,E})\to U_{E/X}(X)\to\textrm{Pic}(X)\overset{f^{*}}{\to}\textrm{Pic}(E). $$ So, you must prove that in your particular case the map $\varGamma(E, \mathbb G_{m,E})\to U_{E/X}(X)$ is surjective. Well, is it?
Regarding the proof: exactness of the above exact sequence boils down to checking that the kernel of $f^{*}\colon \textrm{Pic}(X)\to\textrm{Pic}(E)$ is the same as the kernel of $H^{1}(f^{\flat})$. This follows from the fact that $f^{*}$ factors as $$ H^{1}(X,\mathbb G_{m,X})\to H^{1}(X,f_{*}\mathbb G_{m,E})\hookrightarrow H^{1}(E,\mathbb G_{m,E}), $$ where the first map is $H^{1}(f^{\flat})$ and the second map is the first edge morphism in the Cartan-Leray spectral sequence $H^{ r}(X, R^{ s} f_{*}\mathbb G_{m,E})\Rightarrow H^{ r+s}(E, \mathbb G_{m,E})$.
Observe now the following helpful fact: If $f$ has a section $\sigma\colon X\to E$, then $\mathbb G_{m,E}\to \sigma_{*}\mathbb G_{m,X}$ induces a morphism $f_{*}\mathbb G_{m,E}\to f_{*}\sigma_{*}\mathbb G_{m,X}=\mathbb G_{m,X}$ which splits the sequence $$ 0\to \mathbb G_{m,X}\to f_{*}\mathbb G_{m,E}\to U_{E/X}\to 0. $$ In this case the surjectivity of $\varGamma(E, \mathbb G_{m,E})\to U_{E/X}(X)$ is evident. Does your map have a section?