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$\def\conv{\mathop{\mathrm{conv}}}$Recall that $\conv X$ is the set of all convex combinations of points from $X$. In a convex combination $$ f=\sum_{i=1}^k\alpha_ix_i, \qquad x_i\in X, \quad \alpha_i>0, \quad \sum_{i=1}^k\alpha_i=1, $$ for a point $f\in F$, all the points $x_i$ should lie in $F$ (and hence in $X\cap F$); indeed, if, say, $x_k\notin F$, then $f$ is a relatice interior point of the segment between $x_k$ and $$ \frac1{1-\alpha_k}\sum_{i=1}^{k-1}\alpha_ix_i $$ which is impossible. Thus $f\in\conv(X\cap F)$.