This answer gives the first three Chern classes of the tensor product of any two locally-free sheaves.
I computed the fourth Chern class as $$ c_4(E\otimes F) = \frac{1}{2} c_1^4(E) -5c_1^3(E)c_1(F) -\frac{21}{2}c_1^2(E)c_1^2(F) - 5c_1(E)c_1^3(F) + \frac{1}{2}c_1^4(F) - 4c_1^2(E)c_2(E) - 11c_1(E)c_2(E)c_1(F) - 7c_2(E)c_1^2(F) - 11c_1^2(E)c_2(F) - 11c_1(E)c_1(F)c_2(F) - 4c_1^2(F)c_2(F) + c_2^2(E) + 6c_2(E)c_2(F) + c_2^2(F). $$
where $E,F$ are rank two coherent sheaves (or vector bundles if that is what you prefer).
Is this correct?
Edit: I realised this must be incorrect. Here's the corrected version: $$ -8c_1^3(E)c_1(F) - 16c_1^2(E)c_1^2(F) - 8c_1(E)c_1^3(F) - 6c_1^2(E)c_2(E) - 17c_1(E)c_2(E)c_1(F) - 9c_2(E)c_1^2(F) - 13c_1^2(E)c_2(F) - 17c_1(E)c_1(F)c_2(F) - 6c_1^2(F)c_2(F) - c)2^2(E) + 2c_2(E)c_2(F) - c_2^2(F). $$
Edit: I changed the question to talk about locally-free sheaves.
Let $E$ and $F$ be complex vector bundles of rank $2$ which split as a sum of line bundles: $E \cong E_1\oplus E_2$ and $F\cong F_1\oplus F_2$. Let $x_ i = c_1(E_i)$ and $y_j = c_1(F_j)$. Note that
\begin{align*} c_1(E) &= c_1(E_1\oplus E_2) = c_1(E_1) + c_1(E_2) = x_1 + x_2\\ c_2(E) &= c_2(E_1\oplus E_2) = c_1(E_1)c_1(E_2) = x_1x_2. \end{align*}
Likewise $c_1(F) = y_1 + y_2$ and $c_2(F) = y_1y_2$. So we have the following:
\begin{align*} &\ c_4(E\otimes F)\\ =&\ c_4((E_1\oplus E_2)\otimes(F_1\oplus F_2))\\ =&\ c_4(E_1\otimes F_1\oplus E_1\otimes F_2\oplus E_2\otimes F_1\oplus E_2\otimes F_2)\\ =&\ c_1(E_1\otimes F_1)c_1(E_1\otimes F_2)c_1(E_2\otimes F_1)c_1(E_2\otimes F_2)\\ =&\ [c_1(E_1) + c_1(F_1)][c_1(E_1) + c_1(F_2)][c_1(E_2) + c_1(F_1)][c_1(E_2) + c_1(F_2)]\\ =&\ [x_1 + y_1][x_1 + y_2][x_2 + y_1][x_2 + y_2]\\ =&\ [x_1^2 + x_1y_2 + x_1y_1 + y_1y_2][x_2^2 + x_2y_2 + x_2y_1 + y_1y_2]\\ =&\ [x_1^2 + x_1y_2 + x_1y_1 + c_2(F)][x_2^2 + x_2y_2 + x_2y_1 + c_2(F)]\\ =&\ x_1^2x_2^2 + x_1^2x_2y_2 + x_1^2x_2y_1 + x_1^2c_2(F) + x_1x_2^2y_2 + x_1x_2y_2^2 + x_1x_2y_1y_2 + x_1y_2c_2(F)\\ +&\ x_1x_2^2y_1 + x_1x_2y_1y_2 + x_1x_2y_1^2 + x_1y_1c_2(F) + x_2^2c_2(F) + x_2y_2c_2(F) + x_2y_1c_2(F) + c_2(F)^2\\ =&\ c_2(E)^2 + x_1y_2c_2(E) + x_1y_1c_2(E) + x_1^2c_2(F) + x_2y_2c_2(E) + y_2^2c_2(E) + c_2(E)c_2(F)\\ +&\ x_1y_2c_2(F) + x_2y_1c_2(E) + c_2(E)c_2(F) + y_1^2c_2(E) + x_1y_1c_2(F) + x_2^2c_2(F) + x_2y_2c_2(F)\\ +&\ x_2y_1c_2(F) + c_2(F)^2\\ =&\ c_2(E)^2 + c_2(F)^2 + 2c_2(E)c_2(F) + [x_1y_2 + x_1y_1 + x_2y_2 + y_2^2 + x_2y_1 + y_1^2]c_2(E)\\ +&\ [x_1^2 + x_1y_2 + x_1y_1 + x_2^2 + x_2y_2 + x_2y_1]c_2(F)\\ =&\ c_2(E)^2 + c_2(F)^2 + 2c_2(E)c_2(F) + [(x_1 + x_2)(y_1 + y_2) + (y_1 + y_2)^2 - 2y_1y_2]c_2(E)\\ +&\ [(x_1 + x_2)(y_1 + y_2)^2 + (x_1 + x_2)^2 - 2x_1x_2]c_2(F)\\ =&\ c_2(E)^2 + c_2(F)^2 + 2c_2(E)c_2(F) + [c_1(E)c_1(F) + c_1(F)^2 - 2c_2(F)]c_2(E)\\ +&\ [c_1(E)c_1(F) + c_1(E)^2 - 2c_2(E)]c_2(F)\\ =&\ c_2(E)^2 + c_2(F)^2 - 2c_2(E)c_2(F) + [c_1(E)c_1(F) + c_1(F)^2]c_2(E)\\ +&\ [c_1(E)c_1(F) + c_1(E)^2]c_2(F)\\ =&\ [c_2(E)-c_2(F)]^2 + [c_1(E) + c_1(F)]c_1(F)c_2(E) + [c_1(F) + c_1(E)]c_1(E)c_2(F)\\ =&\ [c_2(E)-c_2(F)]^2 + [c_1(E)+c_1(F)][c_1(F)c_2(E) + c_1(E)c_2(F)]. \end{align*}
By the splitting principle, we see that this identity holds for any rank two complex vector bundles $E$ and $F$. As $E\otimes F$ has rank four, we now have formulae for all of its Chern classes in terms of the Chern classes of $E$ and $F$.
\begin{align*} c_1(E\otimes F) &= 2c_1(E) + 2c_1(F)\\ c_2(E\otimes F) &= 2c_2(E) + 2c_2(F) + c_1(E)^2 + c_1(F)^2 + 3c_1(E)c_1(F)\\ c_3(E\otimes F) &= 6[c_1(E) + c_1(F)][c_2(E) + c_2(F)] + 5c_1(E)^2c_1(F) + 5c_1(E)c_1(F)^2\\ c_4(E\otimes F) &= [c_2(E) - c_2(F)]^2 + [c_1(E) + c_1(F)][c_1(F)c_2(E) + c_1(E)c_2(F)]. \end{align*}