lets say we got PDA $M=<Q, \Sigma,\Gamma,\delta,q_0,+,F>$ ('+' marks the end of the stack) with $L_1 = L_f(M)$.
and we got $A=<P,\Sigma,\delta_1.p_0.F_1>$ with $L_2 = L(A)$.
how can I build a PDA $M' =<Q', \Sigma,\Gamma,\delta,q_0',+,F'>$ with $L_f(M')=L_1\cap L_2$
Not quite sure about what you denote by $L_f(M)$, but I assume that it is the language accepted by a PDA by a final state. I will also assume that the $\delta$ from the $M'$-tuple is meant to be $\delta'$.
The more-or-less standard approach to this problem is to simulate automata $M$ and $A$ simultaneously on the given input. This means that $M'$ should have states of the form $[q,p]$, where $q \in Q$ and $p \in P$. After reading a word $w$, our new automaton $M'$ should be in a state $[q,p]$ if and only if, after reading $w$, $M$ arrives at $q$ and $A$ arrives at $p$. Moreover, the contents of the stack of $M'$ should be the same as for the automaton $M$.
Now, as both automata start their computations from their initial states, the initial state of $M'$ should be $[q_0,p_0]$. For each $c \in \Sigma$, we want to simulate both automata at once, so $\delta'([q,p],c,Z)$ should contain $([r,\delta_1(p,c)],\gamma)$ for all $(r,\gamma) \in \delta(q,c)$. Moreover, $M$ can perform spontantenious $\varepsilon$-transitions – in that case, we need not simulate $A$ at all, so $\delta'([q,p],\varepsilon,Z)$ should contain $([r,p],\gamma)$ for all $(r,\gamma) \in \delta(q,\varepsilon)$. Finally, we want to accept the input if and only if both automata accept, that is $F' = F \times F_1$.
The full construction of $M' = (Q',\Sigma,\Gamma,\delta',q'_0,+,F')$ is as follows:
$Q' = Q \times P$
$\delta'([q,p],c,Z) = \{([r,\delta_1(p,c)],\gamma) ~|~ (r,\gamma) \in \delta(q,c)\}$ for all $q \in Q$, $p \in P$, $c \in \Sigma$, and $Z \in \Gamma$
$\delta'([q,p],\varepsilon,Z) = \{([r,p],\gamma) ~|~ (r,\gamma) \in \delta(q,\varepsilon)\}$ for all $q \in Q$, $p \in P$, and $Z \in \Gamma$
$q'_0 = [q_0,p_0]$
$F' = F \times F_1$
This construction should work regardless the function $\delta_1$ is required to be total or not.