I agree with Henning that your induction should be on $|v|$. Note in the case $|v| = 1$, you're done by definition. It's probably also worth mentioning something about the empty string for completeness.
Let me demonstrate how I would prove this for $|v| = 3$ supposing we knew this were true for when $|v| = 2$. This should illustrate how you should prove $|v| = n$ from assuming $|v| = n - 1$ is valid.
Let $v = v_1v_2v_3$ be a string of length 3. Assume the claim is true for strings of length 2. Then $(uv)^R = (uv_1v_2v_3)^R$. By the recursive definition, $(uv_1v_2v_3)^R = v_3(uv_1v_2)^R$. Note $v_1v_2$ is a string of length 2, so we have $v_3(uv_1v_2)^R = v_3(v_1v_2)^Ru^R$. But by definition, $v_3(v_1v_2)^Ru^R = v_3v_2v_1u^R = (v_1v_2v_3)^Ru^R$.
In almost all inductive proofs, it helps to prove a few small cases by using even smaller cases. If you're still having trouble, try proving this for $|v| = 4$ by using the same ideas as above.