The condition implies directly that $$\text{if }x_1+y_1=x_2+y_2\text{, then }f(x_1)f(y_1)+\alpha x_1y_1=f(x_2)f(y_2)+\alpha x_2y_2~~(\star)$$we get $$\begin{cases} f(t)f(-t)-\alpha t^2=f(0)^2&(1)\\ f(2t)f(-t)-2\alpha t^2=f(t)f(0)&(2)\\ f(-2t)f(t)-2\alpha t^2=f(-t)f(0)&(3)\\ f(2t)f(-2t)-4\alpha t^2=f(0)^2&(4) \end{cases}$$solve $f(2t),f(-2t)$ from $(2),(3)$, plug in $(4)$ while noticing $(1)$, we have $$2\alpha t^2f(0)(2f(0)-f(t)-f(-t))=0.~~~~~(*)$$So two cases: Case 1: $f(0)\neq 0$: So $f(t)+f(-t)=2f(0)$ holds for arbitrary $t\in\mathbb{R}$. Plug in $(1)$ again we have $$f(t)=f(0)\pm t\sqrt{-\alpha}$$implicitly $\alpha<0$. If we have $f(x)=f(0)+x\sqrt{-\alpha}$ and $f(y)=f(0)-y\sqrt{-\alpha}$, directly $f(-x)=f(0)-x\sqrt{-\alpha}$ and $f(-y)=f(0)+y\sqrt{-\alpha}$, consider$$2f(0)^2=f(0)f(x+y)+f(0)f(-x-y)=f(x)f(y)+\alpha xy+f(-x)f(-y)+\alpha xy=2f(0)^2+4\alpha xy$$So $xy=0$. But notice $f(0)=f(0)+0\times\sqrt{-\alpha}=f(0)-0\times\sqrt{-\alpha}$, so the sign can be chosen to be the same for all $x\in\mathbb{R}$, i.e. $$f(x)=f(0)\pm x\sqrt{-\alpha},~\forall x\in\mathbb{R}.$$Plug in the original equation we get $\alpha=-4,f(0)=2$. Case 2: $f(0)=0$: By $(\star)$ we have $$f(x)f(y)+\alpha xy=f(x+y)f(0)+\alpha(x+y)\times0=0,~~\forall x,y\in\mathbb{R}$$which is enough to give $$f(x)=x\sqrt{\alpha},~\forall x\in\mathbb{R}.$$implicitly $\alpha>0$. This function does not satisfy the original condition. To sum up, $$f(x)=2\pm 2x~(\alpha=-4).$$