A morphism $F \to E$ can be thought of as a representation of the quiver $$ A_2 = \{ \bullet \to \bullet \} $$ in the category $Coh(X)$. The category $Rep(A_2,Coh(X))$ is abelian, so what you need is $Ext^1$ in this category. If $S_1$ and $S_2$ denote the simple modules of the first and the second vertices of the quiver, there is an exact sequence $$ 0 \to E \otimes S_2 \to C \to F \otimes S_1 \to 0, $$$$ 0 \to E \otimes S_2 \to C \to F \otimes S_1 \to 0,\qquad(*) $$ where $C = \{F \to E\}$. Using the fact that $Ext^\bullet(S_i,S_i) = \Bbbk$, $Ext^\bullet(S_1,S_2) = \Bbbk[-1]$, $Ext^\bullet(S_2,S_1) = 0$, one gets a long exact sequence $$ 0 \to Hom(C,C) \to Hom(E,E) \oplus Hom(F,F) \to Hom(F,E) \to Ext^1(C,C) \to Ext^1(E,E) \oplus Ext^1(F,F) \to Ext^1(F,E) \to \dots $$ And the space $Ext^1(C,C)$ in its middle is the tangent space to deformations.
EDIT. To get the long exact sequence, note first that applying to $(*)$ the functor $Hom(-,E\otimes S_2)$ one gets $$ 0 \to 0 \to Hom(C,E\otimes S_2) \to Hom(E,E) \to Hom(F,E) \to Ext^1(C,E\otimes S_2) \to Ext^1(E,E) \to Ext^1(F,E) \to \dots $$ and applying to $(*)$ the functor $Hom(-,F \otimes S_1)$ one gets $$ Ext^i(C,F \otimes S_1) \cong Ext^i(F,F). $$ Now, finally, applying to $(*)$ the functor $Hom(C,-)$ and using the above identifications, one gets the desired sequence.