Question on the proof of Hartshorne IV.3.7 Proposition 3.7 in Hartshorne's Algebraic Geometry is the following.

Let $X$ be a curve in $\mathbb P^3$, let $O$ be a point not on $X$,
  and let $\phi: X \rightarrow \mathbb P^2$ be the morphism determined
  by projection from $O$. Then $\phi$ is birational onto its image and
  $\phi(X)$ has at most nodes as singularities, if and only if 
(1) $O$ lies on only finitely many secants of $X$, (2) $O$ is not on
  any tangent line of $X$, (3) $O$ is not on any multisecant of $X$, and
  (4) $O$ is not on any second with coplanar tangent lines.
Proof. Going back to the proof of (II, 7.3), condition (1) says that
  $\phi$ is one-to-one almost everywhere, hence birational. When $O$
  does lie on a secant line, conditions (2), (3), (4) tell us that the
  line meets $X$ in exactly two points P, Q, it is not tangent to X at
  either one, and the tangent lines at P, Q are mapped to distinct lines
  in $\mathbb P^2$. Hence the image $\phi(X)$ has a node at the point.

The proof here is unsatisfying, and I was wondering how to fill in the details. (I looked for a better reference and could not find one.) Geometrically, it's fairly easy to see what's going on. Condition (1) is imposed so that the map is one-to-one almost every where, condition (2) is imposed to make sure the map is injective on tangent vectors, condition (3) is imposed to make sure there are no three-to-one points, and condition (4) is imposed to avoid tacnodes. But it's not clear to me how to translate these into precise scheme-theoretic statements. 
For example, since this is not just a closed immersion, we need to specify a scheme structure on the image. How? And how do we know the image curve is regular at the points where the map is one-to-one?
Further, the modern definition of a node uses the completion of the local ring of the singular point (cf. Vakil's notes, chapter 29). How do we see using this definition that the points in the image that have two pre-images are nodes?
I'd like to do this purely algebraically, in all characteristics. 
 A: By purely algebraically, I will understand that scheme language is allowed. One could make the translation to commutative algebra for all the statements that I make, but somehow that's not the most intuitive way to think about this.
To address your question about the scheme structure: we use the scheme theoretic image, defined in exercise II.3.11 (d). Since $X$ is a curve (in particular it is integral), we see that $Y = \phi(X)$ carries the induced reduced scheme structure. Moreover, by Exercise II.4.4, it is closed in $\mathbb P^2$.
Since the image of an irreducible topological space under any continuous map is irreducible (if you don't know this, you should try to prove this), we conclude that $Y$ is integral. Finally, it has dimension at most $1$ since it is the image of a space of dimension at most $1$, and in fact the dimension has to equal $1$ because otherwise $Y$ would be a point, contradiction assumption (1). Thus, $Y$ is a possibly singular (complete) curve.
Note that the morphism $\phi \colon X \to \mathbb P^2$ is defined by the invertible sheaf $\mathcal O(1)$ and the sections $x_0, x_1, x_2$ (assuming the point $O$ is $[0:0:0:1]$). Let $U \subseteq X$ be the set of $x \in X$ where no secant through $x$ passes through $O$. Note that automatically, no tangent line at $x \in U$ passes through $O$.
Then we can modify the proof of II.7.3: instead of proving that $\phi$ is a closed immersion, we can prove that $\phi$ is a locally closed immersion (closed immersion followed by an open immersion) on $U$. Since $x_3$ is the only nonzero coordinate of $O$, we see that the equation for the line $\overline{OP}$ for any $P \in U$ does not involve the coordinate $x_3$ (if it did, it wouldn't vanish at $O$). This gives criterion (1) of II.7.3: for $P \neq Q$ we take the line $OP$, which by assumption does not pass through $Q$. You can try to work out criterion (2) yourself.
If $V$ denotes the image of $U$ in $Y$, we conclude that the map $U \to V$ is a closed immersion. Since both $U$ and $V$ are integral of dimension $1$, we must have $U \stackrel{\sim}{\to} V$. This answers your second question: $Y$ is regular at $V$ because $X$ is regular and $U \cong V$.
Your last question (about the node) is the most interesting one. But we now leave the realm of this particular morphism $X \to Y$, and we essentially only have the following algebra question:

Question. Let $A \to B$ be a finite ring homomorphism, where $(A,\mathfrak m)$ is local and $B$ has two maximal ideals $\mathfrak n_1, \mathfrak n_2$. Assume $A$ and $B$ are $1$-dimensional rings which are localisations of finite type algebras over an algebraically closed field $k$. Assume that the maps $\Omega_{A/k} \to \Omega_{B_{\mathfrak n_i}/k}$ have distinct kernels. Assume that $A$ is planar and that $B$ is regular. Then $A$ has a node, i.e. $\hat A \cong k[[x,y]]/(xy)$.

I will not answer this question here, but I will give you a hint: because $A$ is planar, you can already write it as $(k[x,y]/(f))_{(x,y)}$ (assume wlog that $\mathfrak m$ is the origin). By Vakil, Exercise 29.B you need to show that $f$ has no degree $0$ or $1$ terms, and that the degree $2$ term is not a perfect square.
To show that a degree $2$ term appears, use the fact that $B$ is regular with two maximal ideals (in Hartshorne's language [Exercise I.5.3], $\mathfrak m$ has multiplicity $2$). Computing a basis for the differentials, in terms of $f$, check that the nonsquare condition follows from the assumption on the map on differentials.
If you also want it to work in characteristic $2$, you need to slightly modify the criterion of Vakil's exercise. This might be a bit tricky.
(Disclaimer: I didn't work out the computation in detail. Please verify that everything works before accepting this answer).
