In Frege proof systems, how to understand 2 if then? For axioms, e.g.
$(p_1 \supset p_3) \supset (p_2 \supset p_3) \supset (p_1 \lor p_2 \supset p_3)$

My understanding is if ($p_1$ can infer $p_3$, and $p_2$ can infer $p_3$), then (if $p_1$ is true or $p_2$ is true, then $p_3$ is true).

If so, how to understand the 2nd $\supset$, why not use $\land$?
The lack of parentheses introduces an ambiguity in the formula.

See e.g. Herbert Enderton, A Mathematical Introduction to Logic, Academic Press (2nd ed., 2001), page 33 on omitting parentheses :


  we adopt the following conventions : [...]
  
  
    
    The negation symbol applies to as little as possible.
    The conjunction and disjunction symbols apply to as little as possible,
    given that convention 2 is to be observed. 
    Where one connective symbol is used repeatedly, grouping is to the right: $α → β → γ$ is $α → (β → γ)$.
    
  


Thus, I'll read the formula as :


  $(P \to Q) \to [(R \to Q) \to ((P \lor R) \to Q)]$.


It is simply Disjunction elimination written as a tautology instead of as a rule.

See Hilbert-style axioms system for intuitionistic logic (but why Frege's ?). 

From it, we can recover the rule using modus ponens twice :


  $(P \to Q), (R \to Q) \vdash (P \lor R) \to Q$.




If we read the formula that way, your interpretation is correct : we can use $\land$ instead to express the same logical principle.

See Exportation :


  $((\varphi \land \psi) \to \sigma) \equiv (\varphi \to (\psi \to \sigma))$.
No.

It's more like:

"If, if p1, then p3, then if, if p2, then p3, then if p1 or p2, then p3."

In your translation you have a conjunction as the antecedent, having two conditionals as the conjuncts of that conjuction.  But, the formula says that the antecedent is a conditional "if p1, then p3".

There does exist a relationship via the exportation and importation laws that Mauro referenced, but such a relationship concerns formulas that differ.

In general, two 'if-thens' can get understood as multiple if-thens.