How to understand this formula in quantum field theory?

The formula holds, because there is a sum over all ${bf p}$. After that sum is performed, ${bf v}cdot{bf v}'$ is essentially the only vector structure that is consistent with rotation symmetry.

To prove the formula, first rewrite it so that the ${bf v}$ and ${bf v}$' are outside the sum: $$sum_{{bf p}}({bf p}cdot{bf v})({bf p}cdot{bf v}')F(p^{2})= sum_{j=1}^{d}v_{j}sum_{j=1}^{d}v_{k}'sum_{{bf p}}p_{j}p_{k}F(p^{2}).$$

Now look just at the final sum, $sum_{{bf p}}p_{j}p_{k}F(p^{2})$. Consider holding the index $j$ fixed (say $j=1$) and varying $k$. If $kneq j$ (say $j=2$) then for every term in the sum with ${bf p}=(p_{j},p_{k},p_{3})$, there is another term in the sum with ${bf p}'=(p_{j},-p_{k},p_{3})$. Since the sum terms $p_{j}p_{k}F(p^{2})$ and $p_{j}'p_{k}'F(p'^{2})$ are equal in magnitude but opposite in sign, they will cancel; and, if fact, every term in the sum will be canceled by a similar one with an inverted $p_{k}rightarrow p_{k}'=-p_{k}$. Thus, the sum is $sum_{{bf p}}p_{j}p_{k}F(p^{2})=0$ if $jneq k$.

The expression $sum_{{bf p}}p_{j}p_{k}F(p^{2})$ transforms like a two-index tensor. (Any product of the components of two vectors ${bf u}$ and ${bf w}$ along with a scalar $phi$—that is, $u_{j}w_{k}phi$—transforms this way. The sum of interest is then just a sum of tensors that transform this way and so transforms this way itself.) The only two-tensor that has components that are nonzero only for $j=k$ is one that is proportional to the identity, with components $delta_{jk}$. [This is essentially just a long-winded way of saying that, since there is no preferred direction, $sum_{{bf p}}p_{1}p_{1}F(p^{2}) =sum_{{bf p}}p_{2}p_{2}F(p^{2})=sum_{{bf p}}p_{3}p_{3}F(p^{2})$.] So we must have $sum_{{bf p}}p_{j}p_{k}F(p^{2})$ be a scalar function times $delta_{jk}$, $$sum_{{bf p}}p_{j}p_{k}F(p^{2})=delta_{jk}G,$$ which leaves us to find $G$.

To get the scalar $G$, we contract the tensor with another $delta_{jk}$, or $$sum_{j=1}^{d}sum_{k=1}^{d}delta_{jk}sum_{{bf p}}p_{j}p_{k}F(p^{2})=sum_{j=1}^{d}sum_{k=1}^{d}delta_{jk}delta_{jk}G,$$ which reduces to $$sum_{{bf p}}p^{2}F(p^{2})=dG,$$ since $sum_{j=1}^{d}sum_{k=1}^{d}delta_{jk}delta_{jk}$ is just the dimensionality of space $d$.

Then it is just a matter of inserting this expression into the earlier ones and working backwards. Solving for $G$, we have $$sum_{{bf p}}p_{j}p_{k}F(p^{2})=delta_{jk}G=delta_{jk}frac{1}{d}sum_{{bf p}}p^{2}F(p^{2}).$$ Then inserting this into the original expression with $v_{j}$ and $v_{k}'$ gives $$sum_{j=1}^{d}v_{j}sum_{j=1}^{d}v_{k}'sum_{{bf p}}p_{j}p_{k}F(p^{2})= sum_{j=1}^{d}v_{j}sum_{j=1}^{d}v_{k}'delta_{jk}frac{1}{d}sum_{{bf p}}p^{2}F(p^{2}),$$ and collecting this in component-free form gives the final expression $$sum_{{bf p}}({bf p}cdot{bf v})({bf p}cdot{bf v}')F(p^{2})=frac{1}{d}{bf v}cdot{bf v}'sum_{{bf p}}p^{2}F(p^{2}).$$