Given a curve $\gamma$ in a Banach space $X$ and a function f defined along the curve s.t. $$\big\Vert f(\gamma(t))f(\gamma(s))\big\Vert\\leq L\big\Vert\gamma(t)\gamma(s)\big\Vert$$ is it possible to extend the Lipschitz functions to the whole of $X$?

It is not always possible to extend when $X$ is a Banach space. Take a Banach space $Y_n$ which contains an $n$ dimensional subspace $E_n$ such that every projection from $Y_n$ onto $E_n$ has norm at least $C_n$ with $C_n\to \infty$. ($Y_n$ can e.g. be $L_1$ and $E_n$ the span of $n$ IID gaussian random variables; then $C_n$ is of order $n^{1/2}$.) Let $X_n = Y_n \oplus_2 E_n$. For the curve in $Y_n$ take any curve in the unit sphere of $E_n \oplus \{0\}$ that contains an $\epsilon_n$ net $A_n$ of the unit sphere of $E_n \oplus \{0\}$. For $f_n$ take the natural isometry from $E_n \oplus \{0\}$ onto $ \{0\} \oplus E_n $ restricted to the curve. Let $F_n$ be an extension of $f_n$ to a Lipschitz mapping on $X_n$; WLOG $F_n$ maps into $ \{0\} \oplus E_n $ since this is a norm one complemented subspace of $X_n$. Let $G_n$ be the positively homogeneous extension of the restriction of $F_n$ to the unit sphere of $X_n$. Then the Lipschitz constant of $G_n$ is at most three times the Lipschitz constant of $F_n$. Compose $G_n$ with the obvious isometry from $ \{0\} \oplus E_n $ onto $E_n \oplus \{0\}$. The restriction of this map to $Y_n$ gives a positively homogenous mapping from $Y_n$ into $E_n$ that is the identity on $A_N$. By the arguments in $$ $$ Johnson, William B.(1OHSN); Lindenstrauss, Joram(ILHEBR) Extensions of Lipschitz mappings into a Hilbert space. Conference in modern analysis and probability (New Haven, Conn., 1982), 189–206, Contemp. Math., 26, Amer. Math. Soc., Providence, RI, 1984 $$ $$ we conclude that if $\epsilon_n$ is sufficiently small, there is a projection from $Y_n$ onto $E_n$ whose norm is no worse than something like ten times the Lipschitz constant of $G_n$. All of this shows that you cannot get Lipschitz extensions with controlled norms. Take an infinite direct sum to get an example where you cannot get any Lipschitz extension. 


If you mean a realvalued function $f$, yes, and keeping the same constant $L$, by a simple construction. Check the last mentioned property listed here. 


The basic extension result for Lipschitz functions is the theorem of Kirszbaum. This works for functions with values in $\mathbb{R}^n$ and is expounded in Federer's book on Geometric Measure Theory. I think that it even works for functions with values in Hilbert space but can't trace a reference. 

