Prove that $P[B_{\tau_2 } > B_{\tau_1 } | B_{\tau_1 } ] = \frac{B_{\tau_1 } - f_2(B_{\tau_1 } , -1 )}{f_2(B_{\tau_1 } ,1 ) - f_2(B_{\tau_1 } , -1 )}$ In what follows $B_t $ denotes a Brownian Motion.

My first question concerns the boject 
$$P[B_{\tau_2 } > B_{\tau_1 }  | B_{\tau_1 } ]$$
What probabilistic object is this referring to? Some regular conditional distribution? 

For the main question the context is that we have a a stopping time $\tau_1 :=  \inf \{t>0 : \ B_t \in \{a, b\} \} $ and a measurable function $f_2: \ \mathbb R \times \{-1 , 1\}  \to \mathbb R$ and have defined [it is assumed that $f_2( \bullet , -1 ) < f_2( \bullet, 1 ) $]
$$\tau_2 := \inf \{t>0 : \ B_t \in \{f_2(B_{\tau_1 }, -1), f_2(B_{\tau_1 } , 1) \} \} $$
Then it is claimed - referring to the Strong Markov property - that

$$P[B_{\tau_2 } > B_{\tau_1 }  | B_{\tau_1 } ] = \frac{B_{\tau_1 } - f_2(B_{\tau_1 } , -1 )}{f_2(B_{\tau_1 } ,1 ) - f_2(B_{\tau_1 } , -1 )}$$

I think with some reference to the (known) fact that for $a < 0 < b $ and $\tau_{a, b } := \inf \{t>0 : B_t \in \{a, b \} \} $
$$P[B_{\tau_{a, b } }  = b  ]= \frac{a}{b-a}$$

How can I prove this?

Any help would be much appreciated!
 A: 
What probabilistic object is this referring to? Some regular conditional distribution? 

"Regularity" corresponds to the property of $\mathrm P(A\mid \mathcal C)$ as a function of $A$ (it should be a measure). Here, you have a lone event. So I the term "regular conditional distribution" is not relevant. 
Per the question, thanks to the strong Markov property of $B$, $\{B_{t+\tau_1} - B_{\tau_1},t\ge 0\}$ is independent of $B_{\tau_1}$ and has the same distribution as $B$. Now we use the following property of conditional expectation (I'm not aware of any name for it):

If $X$ and $Y$ are independent, then, for any nice (say, bounded) jointly measurable function $f(x,y)$, 
  $$
\mathrm E[f(X,Y) \mid Y] = \mathrm E[f(X,y)]\big|_{y=Y}.
$$

Define 
$$
A(y) = \bigl\{x\in C([0,\infty)) \mid \exists t>0:\\ x(t) = f_2 (y,1) - y, \forall s\in [0,t], x(s) > f_2 (y,-1) - y\bigr\} 
$$
Then, 
$$
\mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \mathrm P( B_{\cdot +\tau_1} - B_{\tau_1} \in A(B_{\tau_1}) \mid B_{\tau_1} ).
$$
Using the independence and the above property,
$$
\mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \mathrm P( B_{\cdot +\tau_1} - B_{\cdot} \in A(y) )\big|_{y =B_{\tau_1}} = \mathrm P( B \in A(y) )\big|_{y =B_{\tau_1}}, \tag{1}
$$
where the last equality holds thanks to the strong Markov property.
Now,
$$
\mathrm P( B \in A(y) ) = \mathrm{P} (B_{\tau_{a,b}} = b) = \frac{-a}{b-a}
$$
with $a = f_2(y,-1)-y$, $b = f_2(y,1)-y$ (you have a mistake here, as the probability can't be negative). 
Combining this with (1),
$$
\mathrm P(B_{\tau_2} - B_{\tau_1} > 0 \mid B_{\tau_1} ) = \frac{B_{\tau_1} - f_2(B_{\tau_1},-1)}{f_2(B_{\tau_1},1) - f_2(B_{\tau_1},-1)},
$$
as required.
