APL (Dyalog Extended), 20 bytes

,(×⌿⍤!∘⊂⌹⍥↑⊢⊂⍛!⊢)⍳⍤+ 

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A tacit dyadic function that takes m and n as left and right arguments respectively. Mainly uses the matrix division built-in ⌹ to solve the linear equations:

$$ \begin{bmatrix} \binom{1}{1} & \binom{1}{2} & \cdots & \binom{1}{n+m} \\ \binom{2}{1} & \binom{2}{2} & \cdots & \binom{2}{n+m} \\ \vdots & \vdots & \ddots & \vdots \\ \binom{n+m}{1} & \binom{n+m}{2} & \cdots & \binom{n+m}{n+m} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_{n+m} \end{bmatrix} = \begin{bmatrix} \binom{1}{n} \binom{1}{m} \\ \binom{2}{n} \binom{2}{m} \\ \vdots \\ \binom{n+m}{n} \binom{n+m}{m} \end{bmatrix} \Leftrightarrow Ba = v $$

When \$ B \$ and \$ v \$ are ready, the answer is simply v⌹B. So the main work is to golf the parts to build \$ B \$ and \$ v \$.

How it works

,(×⌿⍤!∘⊂⌹⍥↑⊢⊂⍛!⊢)⍳⍤+ ⍝ Left argument: n, Right argument: m ,(  )⍳⍤+ ⍝ Pass [n m] and [1 .. n+m] to the inner function   ⊢⊂⍛!⊢ ⍝ Compute the nested array version of B ×⌿⍤!∘⊂  ⍝ Compute the nested array version of v   ⍥↑  ⍝ Convert both sides to proper matrix/vector ⌹  ⍝ Solve the linear equation 

Computing \$ B \$

x!y computes \$ \binom{y}{x} \$.

⊢⊂⍛!⊢ ⍝ Left: [n m], Right: [1 .. n+m] ⊢ ⊢ ⍝ Use right argument for both sides (L, R)  ⊂⍛ ⍝ Enclose L, so entire L is mapped to each element of R  ! ⍝ Compute binomial 

The result is a vector of vectors that looks like \$ (\binom{1}{1}, \binom{1}{2}, \cdots, \binom{1}{n+m}), (\binom{2}{1}, \binom{2}{2}, \cdots, \binom{2}{n+m}), \cdots \$. When ↑ is applied, each inner vector becomes a row of the matrix \$ B \$.

Computing \$ v \$

×⌿⍤!∘⊂ ⍝ Left: [n m], Right: [1 .. n+m]   ∘⊂ ⍝ Enclose right ! ⍝ Compute binomial  ⍝ Result is length-2 nested vector [U V] ×⌿⍤  ⍝ Reduce by multiply; compute U×V element-wise 

APL (Dyalog Extended), ₂₀ 18 bytes

,((×⌿!\)⌹⊢!⍨\⊢)⍳⍤+ 

Try it online!

A tacit dyadic function that takes m and n as left and right arguments respectively. Mainly uses the matrix division built-in ⌹ to solve the linear equations:

$$ \begin{bmatrix} \binom{1}{1} & \binom{1}{2} & \cdots & \binom{1}{n+m} \\ \binom{2}{1} & \binom{2}{2} & \cdots & \binom{2}{n+m} \\ \vdots & \vdots & \ddots & \vdots \\ \binom{n+m}{1} & \binom{n+m}{2} & \cdots & \binom{n+m}{n+m} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_{n+m} \end{bmatrix} = \begin{bmatrix} \binom{1}{n} \binom{1}{m} \\ \binom{2}{n} \binom{2}{m} \\ \vdots \\ \binom{n+m}{n} \binom{n+m}{m} \end{bmatrix} \Leftrightarrow Ba = v $$

When \$ B \$ and \$ v \$ are ready, the answer is simply v⌹B. So the main work is to golf the parts to build \$ B \$ and \$ v \$.

How it works

,((×⌿!\)⌹⊢!⍨\⊢)⍳⍤+ ⍝ Left argument: n, Right argument: m ,( )⍳⍤+ ⍝ Pass [n m] and [1 .. n+m] to the inner function ⊢!⍨\⊢ ⍝ Compute B (×⌿!\) ⍝ Compute v ⌹ ⍝ Solve the linear equation 

Computing \$ B \$

x!y computes \$ \binom{y}{x} \$.

⊢!⍨\⊢ ⍝ Left: [n m], Right: [1 .. n+m] ⊢ ⊢ ⍝ Use right argument for both sides (L, R) !⍨\ ⍝ Outer product by flipped binomial:  ⍝ For each pair of l∊L and r∊R, compute r!l or lCr 

Computing \$ v \$

×⌿!\ ⍝ Left: [n m], Right: [1 .. n+m] !\ ⍝ Outer product by binomial function  ⍝ Result is a matrix of two rows [U V] ×⌿ ⍝ Reduce by multiply; compute U×V element-wise