Let $E$ be an affine space attached to a $K$-vector space $T$ and $a \in V\subset E$. Consider the bijection $f:E\rightarrow T$ send $x$ to $x-a$, where $x=(x-a)+a$. Clearly $f(E)$ is a $K$-subspace of $T$.
Suppose for every family $(x_i)_{i\in I}$ of elements of $V$ and every family $(\lambda_i)_{i\in I}$ such that $\sum_{i\in I}\lambda_i=1$, we have $\sum_{i\in I}\lambda_i(x_i-a)+a\in V$. Then $f(V)$ is a $K$-subspace of $T$.
Let $x,y\in V$ and $\lambda,\mu\in K$. We want to show that $\lambda f(x)+\mu f(y)\in f(V)$. There exists $z\in E$ such that $\lambda f(x)+\mu f(y)=f(z)$: i.e. $\lambda(x-a)+\mu(y-a)=z-a$. I want to show that $z\in V$, but I am not able to use the assumption since the sum of the coefficients is not necessarily $1$. Any suggestions?
Note that $a\in V$, so you can use $a$ as one of the $x_i$. Since $a-a=0$, this means that the condition that the sum of the coefficients is $1$ doesn't do anything, since you can change the coefficient of $a-a$ arbitrarily. In your situation, you can rewrite $z-a$ as $$\lambda(x-a)+\mu(y-a)+\nu(a-a)$$ for whatever value of $\nu$ you want, and in particular you can choose $\nu$ such that $\lambda+\mu+\nu=1$.